The property of equality for exponential equations and solving by equating exponents, taking a logarithm of each side, Newton's law of cooling as an exponential model, the property of equality for logarithmic equations, exponentiating each side, and checking every apparent solution for extraneousness.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 7 — Exponential and Logarithmic Functions
Solve Exponential and Logarithmic Equations
Objectives
Five outcomes. Every method here is one function undoing another.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 514-521 — the lesson these objectives are drawn from
Warm-up
Lesson 7.4 made the exponential and the logarithm inverses, and Lesson 7.5 gave you the properties.
Discussion prompt
To solve 2x plus 3 equals 11 you subtract, then divide — each step undoing an operation. What undoes an exponent when the variable is up in the exponent?
Hint: The inverse of an exponential function.
Answer:
A logarithm. It is the only tool that brings a variable down out of an exponent, which is exactly why this lesson comes after the previous two.
\[ 2^x = 5 \;\Longrightarrow\; \log_2 2^x = \log_2 5 \;\Longrightarrow\; x = \log_2 5 \]
Every method in this lesson is the same idea: identify which function has trapped the variable, then apply its inverse to both sides.
Concept
If a variable sits in an exponent, take a logarithm of both sides. If a variable sits inside a logarithm, raise the base to both sides. The two operations undo each other, so one of them always frees the variable.
exponential equation — An equation in which a variable expression occurs as an exponent. A logarithmic equation is one involving a logarithm of a variable expression.
\[ b^x = b^y \;\Longleftrightarrow\; x = y; \qquad \log_b x = \log_b y \;\Longleftrightarrow\; x = y \]
There is one asymmetry. A logarithm only accepts positive arguments, so a logarithmic equation can produce apparent solutions that fail. Every one of them must be checked.
Figure (svg): Two columns comparing the two properties of equality used in this lesson
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 515-517
Section
Section 1
Concept
If b is positive and not 1, then b raised to x equals b raised to y exactly when x equals y. So an exponential equation whose two sides can be written with the same base reduces to an ordinary equation in the exponents.
\[ b^x = b^y \;\Longleftrightarrow\; x = y \]
The work is in the rewriting, not the solving. Both sides must be expressed as powers of one base before the property applies, and the power of a power rule from Lesson 5.1 is what does it.
Figure (svg): An exponential equation solved by rewriting both sides with the same base
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 515-515 — Property of Equality for Exponential Equations
Picture it
Example 1, from a mismatched pair of bases to a linear equation.
Figure (svg): An exponential equation solved by rewriting both sides with the same base
Four and one half are both powers of 2, so the equation collapses to 2x equals negative x plus 3 and the answer is 1.
Worked example
Example 1.
\[ \text{Solve } 4^x = \left(\tfrac{1}{2}\right)^{x-3}. \]
Rewrite both sides with base 2
Why: Four is 2 squared and one half is 2 to the negative 1.
\[ (2 ^{2}) ^{x} = (2 ^{-1}) ^{x - 3} \]
Use the power of a power property
Why: Multiply the exponents on each side.
\[ 2 ^{2 x} = 2 ^{-x + 3} \]
Equate the exponents
Why: The bases match, so the exponents must be equal.
\[ 2 x = -x + 3 \]
Solve the linear equation
Why: Add x to both sides and divide by 3.
\[ x = 1 \]
Figure (svg): An exponential equation solved by rewriting both sides with the same base
\[ x = 1 \]
Verify: substitute into the original
Why: Four to the first is 4. And one half to the power 1 minus 3, which is negative 2, is 2 squared, or 4. Both sides give 4, so the solution checks. Because no domain was restricted anywhere, no extraneous solution can arise from this method.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 515-515
Fill the middle
Example 1, first step.
Fill in the blanks
4^x = \left(\tfrac2___\right)^___ \;\Longrightarrow\; (2^___})^x = (2^___)^___
Why: Four is 2 squared, so the exponent is 2. Recognising every number in the equation as a power of one base is the step that makes the property usable at all.
Worked example
Guided Practice 1 to 3.
\[ \text{Solve } 9^{2x} = 27^{x-1}, \; 100^{7x+1} = 1000^{3x-2}, \; 81^{3-x} = \left(\tfrac{1}{3}\right)^{5x-6}. \]
First: base 3
Why: Nine is 3 squared and 27 is 3 cubed, so 4x equals 3x minus 3.
\[ x = -3 \]
Second: base 10
Why: One hundred is 10 squared and 1000 is 10 cubed, so 14x plus 2 equals 9x minus 6.
\[ 5 x = -8, x = -\frac{8}{5} \]
Third: base 3 again
Why: Eighty-one is 3 to the fourth and one third is 3 to the negative 1.
\[ 12 - 4 x = -5 x + 6 \]
Third: solve
Why: Add 5x to both sides and subtract 12.
\[ x = -6 \]
Figure (svg): The solution to Worked example three more by equating exponents shown as a ladder of expressions, one row per algebraic move
\[ x = -3, \quad x = -\tfrac{8}{5}, \quad x = -6 \]
Verify: check the second in exponent form
Why: At x equal to negative eight fifths, the left exponent is 14 times negative 1.6 plus 2, which is negative 20.4. The right exponent is 9 times negative 1.6 minus 6, also negative 20.4. Checking the exponents is faster than evaluating two enormous powers, and it tests exactly the step that could have gone wrong.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 515-515
Trap
\[ 4^x = 8^{x-1} \]
Set the exponents equal straight away
Why: The property is applied before checking that the bases match.
\[ x = x - 1 \quad \text{(wrong, and impossible)} \]
The property requires the SAME base on both sides. Four and 8 are not the same base, even though both are powers of 2.
\[ (2^2)^x = (2^3)^{x-1} \;\Longrightarrow\; 2^{2x} = 2^{3x-3} \]
Rewrite with a common base first, then equate
Why: Only after both sides read as a power of 2 does the property apply.
\[ 2x = 3x - 3 \;\Longrightarrow\; x = 3 \]
Check: 4 cubed is 64, and 8 squared is 64. Rewriting is the whole method; equating is the easy line at the end.
Sorting
Some equations rewrite easily and some do not.
Sort into buckets
Sort each equation by the method you would choose.
Scanning both sides for a shared base takes a few seconds and decides which of the two methods to use. When in doubt, taking a logarithm always works.
Prediction
Commit before reasoning.
Predict first
Why does b to the x equal b to the y force x to equal y?
Correct: Because an exponential function is one-to-one, so it never repeats an output.
\[ b^x \text{ is one-to-one for } b > 0, \; b \neq 1 \]
Why: Lesson 7.1's graphs are strictly rising for a base above 1 and strictly falling for a base below 1, so each output is produced by exactly one input. That is what being one-to-one means, and it is also why the exponential has an inverse at all. The condition that b is not 1 matters for the same reason: 1 to any power is 1, so that function repeats every output.
Ranking
Solving by equating exponents.
Put in order
Why: Step one is a decision, not a computation, and it is the one that fails most often — if no shared base exists, the whole route is closed and taking a logarithm is the alternative. Steps two and three are Lesson 5.1 material doing familiar work.
Section
Section 2
Concept
When the two sides cannot conveniently share a base, take a logarithm of each side. The inverse property brings the variable down out of the exponent, and the change-of-base formula turns the answer into a number.
\[ 4^x = 11 \;\Longrightarrow\; x = \log_4 11 = \frac{\log 11}{\log 4} \approx 1.73 \]
Any base of logarithm works, since the change-of-base formula gives the same value either way. Taking the logarithm to the equation's own base is tidiest, and taking a common or natural logarithm gets to the calculator fastest.
Figure (svg): An exponential equation solved by taking a logarithm of both sides
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 516-516 — Take a logarithm of each side
Picture it
Example 2, one inverse property and one change of base.
Figure (svg): An exponential equation solved by taking a logarithm of both sides
Four to the 1.73 is about 11, so the answer checks. Bracketing agrees too: 4 squared is 16, so the exponent must be below 2.
Worked example
Example 2.
\[ \text{Solve } 4^x = 11. \]
Take log base 4 of each side
Why: Applying the same function to equal quantities keeps them equal.
\[ \log _{4}(4 ^{x}) = \log _{4} 11 \]
Use the inverse property
Why: The logarithm of the base raised to a power returns that power.
\[ x = \log _{4} 11 \]
Change the base
Why: Divide the common logarithm of 11 by that of 4.
\[ x = \log 11 / \log 4 \]
Compute
Why: One point oh four one four over 0.6021.
\[ \text{about } 1.73 \]
Figure (svg): An exponential equation solved by taking a logarithm of both sides
\[ x \approx 1.73 \]
Verify: bracket the answer
Why: Four to the first is 4 and 4 squared is 16, and 11 lies between them, so the exponent must be between 1 and 2 — and 1.73 is, nearer 2 because 11 is nearer 16. Bracketing before computing catches an inverted change-of-base fraction, which would have given 0.578 instead.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 516-516
Fill the middle
Example 2.
Fill in the blanks
\log_4(4^x) = \log_4 11 \;\Longrightarrow\; x = \log_4 11
Why: The logarithm base 4 of 4 to the x is simply x, by the inverse property of Lesson 7.4. That single step is the entire point of taking a logarithm: it moves the variable from the exponent down to ground level.
Worked example
Guided Practice 4 to 6.
\[ \text{Solve } 2^x = 5, \; 7^{9x} = 15, \; 4e^{-0.3x} - 7 = 13. \]
First: take a logarithm
Why: The natural logarithm of 5 over that of 2.
\[ x = 2.322 \]
Second: bring the whole exponent down
Why: Nine x equals the logarithm base 7 of 15, which is about 1.3917.
\[ 9 x = 1.3917 \]
Second: divide by 9
Why: The coefficient on x is cleared last.
\[ x = 0.155 \]
Third: isolate the power first
Why: Add 7, divide by 4, then take the natural logarithm.
\[ e ^{-0.3 x} = 5, x = -5.365 \]
Figure (svg): The solution to Worked example three more by taking a logarithm shown as a ladder of expressions, one row per algebraic move
\[ x \approx 2.322, \; 0.155, \; -5.365 \]
Verify: check the third by substituting
Why: At x equal to negative 5.365 the exponent is 1.6094, and e to that is 5. Then 4 times 5 minus 7 is 13, which matches. Notice the order: the power had to be isolated on one side BEFORE the logarithm was taken, exactly as a square root must be isolated before squaring.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 516-516
Error analysis
A student solves an equation with a coefficient and a constant attached to the power.
Annotate
On: \( 4e^{-0.3x} - 7 = 13 \;\Longrightarrow\; \ln(4e^{-0.3x}) - \ln 7 = \ln 13 \)
The rule is the same one that governs radicals: isolate the trapped part on one side before applying the inverse operation to both sides.
Ranking
Solving 4e to the negative 0.3x minus 7 equals 13.
Put in order
Why: The first two steps strip everything that is not the power, which is the reverse order of operations applied to the outside of the expression. Only once e to the power stands alone does the logarithm have something clean to undo.
Matching
Bracket each answer before matching.
Match the pairs
Why: The third is small because the 9 multiplying x has to be divided out at the end, and the fourth is negative because a negative rate was in the exponent while the power had to exceed 1. Both signs are predictable before any arithmetic.
Prediction
Commit before reasoning.
Predict first
Example 2 took log base 4. What if you take the natural logarithm instead?
Correct: You get the same answer, since ln 11 over ln 4 equals log 11 over log 4.
\[ x\ln 4 = \ln 11 \;\Longrightarrow\; x = \frac{\ln 11}{\ln 4} \approx 1.73 \]
Why: Taking the natural logarithm gives x times ln 4 equal to ln 11, so x is ln 11 over ln 4, which is 2.3979 over 1.3863, or 1.73 — the same value. This is the change-of-base formula seen from the solving side: the base you take is a matter of convenience, never of correctness. Taking a common or natural logarithm is usually quickest because those are the calculator's keys.
Section
Section 3
Concept
A cooling substance starting at one temperature in surroundings at another approaches the surrounding temperature exponentially. Solving for the time takes exactly the method of the previous idea, once the surrounding temperature has been subtracted off.
\[ T = (T_0 - T_R)e^{-rt} + T_R \]
The surrounding temperature plays the role of the horizontal asymptote from Lesson 7.2. Subtracting it is what leaves a pure exponential for the logarithm to undo.
Figure (svg): Newton's law of cooling applied to an overheated car engine
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 516-516 — Newton's law of cooling
Picture it
Example 3: from 280 degrees toward 80, with a cooling rate of 0.0048.
Figure (svg): Newton's law of cooling applied to an overheated car engine
The curve flattens toward 80 and never reaches it. Two hundred thirty degrees is crossed at about 60 minutes, which answers the question asked.
Worked example
Example 3.
\[ \text{With } T = (T_0-T_R)e^{-rt}+T_R, \; T_0=280, \; T_R=80, \; r=0.0048, \text{ find } t \text{ when } T=230. \]
Substitute every known value
Why: The starting temperature minus the air temperature is 200.
\[ 230 = 200 e ^{-0.0048 t} + 80 \]
Subtract the surrounding temperature
Why: This strips the constant that the curve approaches.
\[ 150 = 200 e ^{-0.0048 t} \]
Divide by the coefficient
Why: One hundred fifty over 200 is 0.75.
\[ 0.75 = e ^{-0.0048 t} \]
Take the natural logarithm and divide
Why: The natural logarithm of 0.75 is about negative 0.2877.
\[ t\text{ about } 60 \]
Figure (svg): Newton's law of cooling applied to an overheated car engine
\[ t \approx 60 \text{ minutes} \]
Verify: substitute the answer back
Why: At t equal to 60 the exponent is negative 0.288, and e to that is 0.7498. Then 200 times 0.7498 is 150, plus 80 gives 230 — the temperature asked for. And the answer is sensible: an hour of waiting matches everyday experience of an overheated engine.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 516-516
Fill the middle
Example 3, second step.
Fill in the blanks
230 = 200e^150 + 80 \;\Longrightarrow\; ___ = 200e^___
Why: Subtracting 80 leaves 150, the amount by which the engine is still above the air. That difference is what decays exponentially; the total temperature does not.
Worked example
The same law, asked the other way round.
\[ \text{With the same model, find } T \text{ at } t = 30 \text{ and the time to reach } 200\text{ degrees.} \]
Evaluate at 30 minutes
Why: The exponent is negative 0.144, and e to that is about 0.8659.
\[ 200(0.8659) + 80 = 253 \]
Set up the second question
Why: Two hundred degrees means 120 above the surrounding air.
\[ 120 = 200 e ^{-0.0048 t} \]
Divide and take the natural logarithm
Why: Point six has natural logarithm about negative 0.5108.
\[ -0.5108 = -0.0048 t \]
Divide by the rate
Why: Point five one zero eight over 0.0048.
\[ \text{about } 106\text{ minutes} \]
Figure (svg): The solution to Worked example a second question on the same model shown as a ladder of expressions, one row per algebraic move
\[ 253\degree \text{ at } t=30; \quad t \approx 106 \text{ min} \]
Verify: check the two answers against each other
Why: At 30 minutes it is 253, at 60 minutes 230, and at 106 minutes 200. The drops are 27, then 23, then 30 over a longer stretch — slowing down, as an approach to an asymptote must. A model that cooled at a steady rate would have crossed 80 and kept going, which no engine does.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 516-516
Trap
\[ 230 = 200e^{-0.0048t} + 80 \]
Take the natural logarithm of each side immediately
Why: The exponential is visible, so the logarithm is applied at once.
\[ \ln 230 = \ln(200e^{-0.0048t}) + \ln 80 \quad \text{(wrong)} \]
The logarithm of a sum is not the sum of the logarithms. No property of Lesson 7.5 permits that line.
\[ 150 = 200e^{-0.0048t} \;\Longrightarrow\; 0.75 = e^{-0.0048t} \]
Strip the constant and the coefficient first
Why: The exponential must stand alone on one side before the logarithm is applied.
\[ \ln 0.75 = -0.0048t \;\Longrightarrow\; t \approx 60 \]
The constant being stripped is the horizontal asymptote from Lesson 7.2. Every model of this shape needs it removed before anything else can happen.
Prediction
Commit before reasoning.
Predict first
According to the model, when does the engine's temperature equal the air temperature of 80?
Correct: Never — 80 is a horizontal asymptote the curve only approaches.
\[ 80 = 200e^{-0.0048t} + 80 \;\Longrightarrow\; e^{-0.0048t} = 0, \text{ impossible} \]
Why: Setting T equal to 80 gives 0 equal to 200 times e to a power, and no real exponent makes an exponential zero. Algebraically the equation has no solution; physically the model says the engine gets arbitrarily close without ever quite matching the air. This is the same asymptote behaviour as Lesson 7.2's decay curves, and it is why cooling questions always ask about a temperature strictly above the surroundings.
Ranking
Solving a cooling model for time.
Put in order
Why: Steps two and three are the isolation phase, working outward-in on everything wrapped around the exponential. Taking the logarithm before either of them is the single most common error in modelling problems of this kind.
Comparison
Fill the blanks. Each letter has a job.
Comparison matrix
| Piece | Meaning | Value here |
|---|---|---|
| T_0 | starting temperature | 280 degrees |
| T_R | surrounding temperature, the asymptote | 80 degrees |
| r | cooling rate | 0.0048 |
| T_0 - T_R | the gap that decays | 200 degrees |
Reading the model before touching it turns a four-letter formula into a description of a physical situation, and that reading is what tells you which piece to strip first.
Section
Section 4
Concept
If two logarithms with the same base are equal, their arguments are equal. If a logarithm equals a number, raise the base to each side and the inverse property frees the argument.
\[ \log_b x = \log_b y \;\Longleftrightarrow\; x = y; \qquad \log_b(u) = k \;\Longrightarrow\; u = b^k \]
Both routes are the same principle as the exponential case, run in the other direction. Which one to use is decided by what stands on the right side: a logarithm, or a plain number.
Figure (svg): A logarithmic equation solved by raising the base to each side
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 517-517 — Property of Equality for Logarithmic Equations
Picture it
Example 5: a logarithm equal to 3, freed by raising 4 to each side.
Figure (svg): A logarithmic equation solved by raising the base to each side
Four cubed is 64, so the argument is 64 and the rest is linear. The check confirms it: the logarithm base 4 of 64 is indeed 3.
Worked example
Example 4 and Guided Practice 7.
\[ \text{Solve } \log_5(4x-7) = \log_5(x+5) \text{ and } \ln(7x-4) = \ln(2x+11). \]
First: equate the arguments
Why: The bases match, so the arguments must be equal.
\[ 4 x - 7 = x + 5 \]
First: solve
Why: Subtract x, add 7, divide by 3.
\[ x = 4 \]
Second: equate the arguments
Why: Both sides are natural logarithms, so the same property applies.
\[ 7 x - 4 = 2 x + 11 \]
Second: solve
Why: Subtract 2x, add 4, divide by 5.
\[ x = 3 \]
Figure (svg): The solution to Worked example equate the arguments shown as a ladder of expressions, one row per algebraic move
\[ x = 4 \quad \text{and} \quad x = 3 \]
Verify: substitute both back
Why: For the first, 4 times 4 minus 7 is 9 and 4 plus 5 is 9, so both sides read the logarithm base 5 of 9. For the second, 7 times 3 minus 4 is 17 and 2 times 3 plus 11 is 17. Both arguments are positive in each case, so neither solution is extraneous.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 517-517
Sorting
Look at the right side of the equation.
Sort into buckets
Sort each logarithmic equation.
Reading the right side first decides the whole route, and reading the left side tells you whether a condensing step has to come before it.
Worked example
Example 5 and Guided Practice 8.
\[ \text{Solve } \log_4(5x-1) = 3 \text{ and } \log_2(x-6) = 5. \]
First: exponentiate using base 4
Why: Raising 4 to each side frees the argument.
\[ 5 x - 1 = 4 ^{3} = 64 \]
First: solve
Why: Add 1 and divide by 5.
\[ x = 13 \]
Second: exponentiate using base 2
Why: Two to the fifth is 32.
\[ x - 6 = 32 \]
Second: solve
Why: Add 6 to each side.
\[ x = 38 \]
Figure (svg): A logarithmic equation solved by raising the base to each side
\[ x = 13 \quad \text{and} \quad x = 38 \]
Verify: check both arguments
Why: For the first, 5 times 13 minus 1 is 64, and the logarithm base 4 of 64 is 3 because 4 cubed is 64. For the second, 38 minus 6 is 32, and the logarithm base 2 of 32 is 5. Both arguments came out positive, which is the condition an answer must satisfy to be genuine.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 517-517
Trap
\[ \log_4(5x-1) = 3 \]
Raise 10 to each side
Why: The common logarithm's base is used out of habit.
\[ 10^{\log_4(5x-1)} = 1000 \quad \text{(wrong)} \]
The left side does not simplify at all: the inverse property needs the exponential's base to match the logarithm's base.
\[ 4^{\log_4(5x-1)} = 4^3 \]
Raise the logarithm's own base to each side
Why: Only matching bases cancel, as Lesson 7.4's inverse property requires.
\[ 5x - 1 = 64 \;\Longrightarrow\; x = 13 \]
Read the subscript before choosing the base to exponentiate with. A logarithm with no subscript is base 10, and ln is base e.
Fill the middle
Example 5.
Fill in the blanks
\log_4(5x-1) = 3 \;\Longrightarrow\; 5x - 1 = 4^3}
Why: The right side of the equation becomes the exponent on the base, giving 64. This is simply the definition of a logarithm from Lesson 7.4 read as a conversion between forms.
Matching
Two methods, four equations.
Match the pairs
Why: The first two reduced to linear equations at once; the last two needed one exponentiation each before becoming linear. In every case the final equation was one you could already solve in Lesson 1.3.
Prediction
Commit before reasoning.
Predict first
Why may you raise the base to each side of an equation and keep it true?
Correct: Because applying the same one-to-one function to equal quantities keeps them equal.
\[ u = v \;\Longrightarrow\; b^u = b^v, \text{ for any base } b \]
Why: If two quantities are equal, then feeding both into the same function gives equal outputs — that alone justifies the step. Being one-to-one matters for the reverse direction: it guarantees no new solutions are introduced. Squaring both sides, by contrast, is not one-to-one, which is exactly why Lesson 6.6's radical equations could gain false solutions and these cannot from this step.
Section
Section 5
Concept
When two logarithms sit on the same side, condense them into one before exponentiating. Condensing widens the domain, so every apparent solution must be substituted back and any that makes an argument zero or negative discarded.
\[ \log 8x + \log(x-20) = 3 \;\Longrightarrow\; 8x(x-20) = 1000 \]
This is the same danger as squaring both sides of a radical equation in Lesson 6.6, and it has the same cure: the check is part of the method, not an optional extra.
Figure (svg): Two apparent solutions tested against the original equation, one surviving
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 518-518 — Extraneous solutions
Picture it
Example 6: the quadratic gives 25 and negative 5.
Figure (svg): Two apparent solutions tested against the original equation, one surviving
At 25 both arguments are positive and the equation holds. At negative 5 both arguments are negative, so neither logarithm exists at all.
Worked example
Example 6, a multiple-choice item.
\[ \text{Solve } \log 8x + \log(x-20) = 3. \]
Condense with the product property
Why: Two logarithms on one side become one logarithm of a product.
\[ \log [8 x(x - 20)] = 3 \]
Exponentiate using base 10
Why: The inverse property frees the argument.
\[ 8 x(x - 20) = 1000 \]
Write in standard form and divide
Why: Eight x squared minus 160x minus 1000, all over 8.
\[ x ^{2} - 20 x - 125 = 0 \]
Factor and check both roots
Why: The factors give 25 and negative 5; only 25 keeps both arguments positive.
\[ x = 25 \]
Figure (svg): Two apparent solutions tested against the original equation, one surviving
\[ x = 25 \]
Verify: substitute both apparent solutions
Why: At 25 the equation reads the logarithm of 200 plus the logarithm of 5, which is the logarithm of 1000, or 3 — correct. At negative 5 the arguments are negative 40 and negative 25, and no logarithm of a negative number exists, so that root is extraneous. A graph agrees: the two curves cross only once.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 518-518
Two truths and a lie
All three are about extraneous solutions.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. What matters is whether each ARGUMENT is positive, not whether x is. In the equation log base 2 of x plus 6 equals 5, the solution x equals 26 is positive, but negative solutions are perfectly possible elsewhere: log of x plus 100 equals 2 gives x equal to negative 90, and the argument 10 is positive, so it is genuine. Test the arguments, never the sign of x.
Worked example
Guided Practice 9 and 10.
\[ \text{Solve } \log 5x + \log(x-1) = 2 \text{ and } \log_4(x+12) + \log_4 x = 3. \]
First: condense and exponentiate
Why: Five x times x minus 1 equals 10 squared.
\[ 5 x ^{2} - 5 x = 100 \]
First: solve the quadratic and check
Why: Dividing by 5 gives x squared minus x minus 20, which factors.
\[ x = 5\text{ or } x = -4 \]
Second: condense and exponentiate
Why: X times x plus 12 equals 4 cubed.
\[ x ^{2} + 12 x - 64 = 0 \]
Second: solve and check
Why: The factors give 4 and negative 16; only 4 keeps both arguments positive.
\[ x = 4 \]
Figure (svg): The solution to Worked example two more with a check shown as a ladder of expressions, one row per algebraic move
\[ x = 5 \quad \text{and} \quad x = 4 \]
Verify: check the surviving answers
Why: For the first, the logarithm of 25 plus the logarithm of 4 is the logarithm of 100, which is 2. For the second, the logarithm base 4 of 16 plus the logarithm base 4 of 4 is 2 plus 1, or 3. Both check. The discarded roots, negative 4 and negative 16, each made an argument negative.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 518-518
Error analysis
A student solves an equation with two logarithms on the left and reports both answers.
Annotate
On: \( \log 8x + \log(x-20) = 3 \;\Longrightarrow\; x = 25 \text{ or } x = -5 \)
Every root of the quadratic has to be tested in the ORIGINAL equation, not in the condensed one. The condensed form is a consequence, not an equivalent.
Fill the middle
Example 6, first step.
Fill in the blanks
\log 8x + \log(x-20) = \log[8x(x-20)] = 3
Why: The product property joins the two logarithms into one logarithm of the product. Only then does the equation have a single logarithm that exponentiating can free.
Sorting
Test the arguments, not the sign of x.
Sort into buckets
Sort each apparent solution.
In each of these the extraneous root satisfied the quadratic perfectly. Only the original equation can tell the difference, which is why the check is part of the method.
Prediction
Commit before reasoning.
Predict first
Why does the quadratic have a root the original equation does not?
Correct: Because condensing replaced two positivity conditions with one on their product.
\[ \text{original needs } x > 20; \quad \text{condensed allows } x < 0 \text{ too} \]
Why: The original equation requires 8x positive AND x minus 20 positive, which forces x above 20. The condensed form only requires their product to be positive, which is also satisfied when BOTH are negative — and that is precisely the region where the false root lives. The condensed equation asks less than the original, so it can have extra solutions. Squaring both sides in Lesson 6.6 loosened the same way.
Comparison
Fill the blanks. What you see decides what you do.
Comparison matrix
| What you see | Method | Watch out for |
|---|---|---|
| Powers with a shared base | equate the exponents | the bases must match first |
| A power equal to a plain number | take a logarithm of each side | isolate the power first |
| A logarithm equal to a logarithm | equate the arguments | check that both stay positive |
| A logarithm equal to a number | exponentiate each side | use the logarithm's own base |
The fifth case, two logarithms on one side, is the fourth with a condensing step in front — and it is the only one that regularly produces extraneous solutions.
Pattern
One decision, then one of four routes.
Exponential equations cannot produce extraneous solutions, because an exponential accepts every real input. Logarithmic ones regularly do.
OpenStax Algebra and Trigonometry 2e, §6.6 Exponential and Logarithmic Equations §6.6
Check
Equating exponents. Rewrite first.
Check your understanding
Solve 9^(2x) = 27^(x - 1).
Answer: A
Why: Both sides become powers of 3, giving 4x = 3x - 3, so x = -3.
Check
Exponentiating. Which base?
Check your understanding
Solve log_2(x - 6) = 5.
Answer: A
Why: Raising 2 to each side gives x - 6 = 32, so x = 38.
Check
Extraneous solutions. Check the arguments.
Check your understanding
Solve log 8x + log(x - 20) = 3.
Answer: A
Why: The quadratic gives 25 and -5, but -5 makes both arguments negative.
Real world
Carbon-14 decays so that the fraction remaining after t years is e raised to negative 0.000121 times t.
Discussion prompt
A bone fragment retains 22 percent of its original carbon-14. Estimate its age, and explain which step required a logarithm.
Hint: Set the fraction equal to 0.22.
Answer:
\[ 0.22 = e^{-0.000121t} \]
\[ \ln 0.22 = -0.000121t \;\Longrightarrow\; -1.5141 = -0.000121t \]
\[ t = \frac{1.5141}{0.000121} \approx 12{,}513 \text{ years} \]
The fragment is roughly 12,500 years old. The logarithm was needed at exactly one point: the unknown sat in the exponent, and nothing but the inverse of the exponential could bring it down.
This is the same three-line shape as the cooling problem — isolate the power, take the natural logarithm, divide by the rate. Radiocarbon dating, drug half-lives, capacitor discharge and compound interest all reduce to it, which is why this one method is worth being fluent in rather than merely able to reproduce.
Commit first
Answer, then rate your confidence honestly.
Predict first
Can an exponential equation like 4 to the x equals 11 have an extraneous solution?
Correct: No — an exponential accepts every real input, so nothing gets excluded.
\[ \text{domain of } b^x: \text{ all reals}; \quad \text{domain of } \log_b x: \; x > 0 \]
Why: The domain of an exponential function is all real numbers, so no candidate solution can fail on domain grounds. Logarithmic equations are the opposite: a logarithm accepts only positive arguments, and condensing quietly relaxes that restriction, which is where false roots come from. Checking an exponential answer is still worthwhile as arithmetic insurance, but it can never uncover an extraneous solution the way checking a logarithmic one can.
Explain it
They can solve 2x plus 3 equals 11 and have just met logarithms.
Discussion prompt
In four sentences or fewer, explain how to decide what to do to both sides of 3 to the x equals 20.
Hint: Ask what is holding the variable.
Answer:
Look at where the x is. Here it is up in the exponent, and the only thing that brings an exponent down is a logarithm.
So take a logarithm of both sides, exactly as you would subtract from both sides to move a constant. That gives x equal to the logarithm base 3 of 20, which a calculator turns into about 2.727 by dividing the logarithm of 20 by the logarithm of 3.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For common bases, list the powers of 2 and of 3 up to a few hundred and look for every number in the equation there. For isolating, strip everything outside the power in reverse order of operations. For exponentiating, read the subscript and use that base. For extraneous roots, substitute each answer and check every argument's sign — never the sign of x itself.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a decision page for solving. Across the top, write the four cases: powers with a shared base, a power equal to a number, a logarithm equal to a logarithm, and a logarithm equal to a number. Under each, write the method in one line and work one example of your own from start to finish. Down the left margin, write the two properties of equality, one for exponential equations and one for logarithmic, and beside each note whether extraneous solutions are possible and why. In the middle, work Example 6 in full and mark clearly the line where the domain got wider. Bottom right, write the cooling model and the three-step isolation it needs, then use it to answer one question you invent. Finish with a boxed rule in your own words for when a check is mandatory rather than optional.
If your boxed rule says to check whenever the answer looks odd, tighten it: the check is mandatory for every logarithmic equation, and the thing to test is each argument's sign.
Recap
Five things, and every one of them is a function undoing another.
| If you see | Then |
|---|---|
| Powers with a shared base | Equate the exponents |
| A power equal to a number | Isolate it, then take a logarithm |
| A logarithm equal to a logarithm | Equate the arguments |
| A logarithm equal to a number | Exponentiate with that logarithm's base |
| Two logarithms on one side | Condense first |
| Any logarithmic equation solved | Substitute back and check every argument |
Lesson 7.7 fits exponential and power models to data, using the properties of logarithms to turn a curved scatter plot into a straight line.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.6 Solve Exponential and Logarithmic Equations §7.6, pp. 514-521 — everything on these slides traces back here
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