Puts the chapter's machinery to work. Solves exponential equations by taking a logarithm of both sides so the power property can bring the exponent down, or by equating exponents when a common base is available. Solves logarithmic equations by condensing and converting to exponential form, and insists on checking, since condensing can produce solutions the original equation rejects.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 4 — Exponential and Logarithmic Functions
§4.6 Exponential and Logarithmic Equations, pp. 553-568
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 553-568 — the pages these objectives are drawn from
Warm-up
Two families, and telling them apart takes one glance.
Discussion prompt
Look at 3 to the power x equals 20, and at the logarithm of x plus 2 equals 3. What is different about where the unknown is?
Hint: In each, is x inside an exponent or inside a logarithm?
Answer:
In the first, x is in the exponent, so it needs an operation that brings exponents down — a logarithm.
In the second, x is inside a logarithm, so it needs the opposite: converting to exponential form to get it out.
So the two families need opposite moves, and identifying which you have is the first step in every problem here. There is also a difference in the checking: the second family can produce answers that must be thrown away, and the first cannot.
Concept
If the unknown is in an exponent, take a logarithm to bring it down. If it is inside a logarithm, exponentiate to bring it out. Each family is solved by the other's function.
\[ b^x=y \;\Longrightarrow\; x=\log_b y, \qquad \log_b x=y \;\Longrightarrow\; x=b^y \]
Both moves are legitimate because both functions are one-to-one, so applying either to both sides of an equation preserves its solutions. The catch is on the logarithmic side, where condensing before converting can enlarge the domain and admit solutions the original never had.
Figure (svg): Two solving strategies side by side: taking a logarithm of both sides when the unknown is in an exponent, and converting to exponential form when the unknown is inside a logarithm
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 553-558
Section
Section 1
Concept
When both sides of an exponential equation can be written as powers of the same base, the exponents must be equal. No logarithm is needed.
The one-to-one property is what licenses the step. If two powers of the same base are equal, their exponents must be equal — because the exponential function never repeats a value, which is §1.7's horizontal line test doing real work here.
Figure (svg): The one-to-one property shown as a shortcut: when both sides can be written as powers of the same base, the exponents may simply be equated
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 553-557
Picture it
When a common base is available, the whole problem is one rewriting.
Figure (svg): The one-to-one property shown as a shortcut: when both sides can be written as powers of the same base, the exponents may simply be equated
No logarithms appeared anywhere. The answer is exact and required no calculator, which is why this route is worth checking for first.
Worked example
Recognise both sides as powers of the same number.
\[ \text{Solve } 9^{x-1}=27. \]
Find a common base
Why: Both 9 and 27 are powers of 3.
\[ \text{base } 3 \]
Rewrite both sides
Why: Nine is 3 squared; 27 is 3 cubed.
\[ 3 ^{2(x - 1)} = 3 ^{3} \]
Equate the exponents
Why: Valid since exponentials are one-to-one.
\[ 2(x - 1) = 3 \]
Solve
Why: Expand and isolate.
\[ x = \frac{5}{2} \]
Figure (svg): The one-to-one property shown as a shortcut: when both sides can be written as powers of the same base, the exponents may simply be equated
\[ x=\tfrac{5}{2} \]
Verify: substitute back
Why: Nine to the power three halves is the cube of the square root of 9, which is 27 — correct. Note the exponent on the left had to be multiplied by 2 when 9 was rewritten as 3 squared, which is the step most often dropped.
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 554-556
Sorting
Both sides must be recognisable powers of one number.
Sort into buckets
Sort each equation.
Worked example
Reciprocals are negative powers.
\[ \text{Solve } 2^x=\tfrac{1}{32}. \]
Recognise the right side
Why: Thirty two is 2 to the fifth.
\[ \frac{1}{32} = 1 / 2 ^{5} \]
Write it as a negative power
Why: A reciprocal is a negative exponent.
\[ = 2 ^{-5} \]
Equate the exponents
Why: Both sides are powers of 2.
\[ x = -5 \]
Check
Why: Two to the negative 5 is one thirty-second.
Figure (svg): The solution to Worked example a fractional common base shown as a ladder of expressions, one row per legal move
\[ x=-5 \]
Verify: notice why no logarithm was needed
Why: The right-hand side happened to be an exact power of 2, so the exponents could be matched directly. Had it been one thirtieth instead, no common base would exist and a logarithm would be required — which is the test for whether this technique applies.
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 556-557
Trap
\[ 9^{x-1}=3^3 \;\Longrightarrow\; x-1=3 \]
Rewrite 27 as a power of 3 and equate
Why: The right side is correctly written as 3 cubed.
The left side is left as 9 to the x minus 1, and its exponent is equated directly.
The left side must be rewritten over base 3 too. Nine is 3 squared, so 9 to the x minus 1 is 3 to the power 2 times the quantity x minus 1.
Equating gives 2 times the quantity x minus 1 equal to 3, so x is five halves rather than 4.
Both sides must share the base before the exponents can be equated. Rewriting a base multiplies the existing exponent, and dropping that multiplication is the standard error here.
Faded example
Solve 25 to the power x equals 125.
Fill in the blanks
5^2x} = 5^3 \;\Longrightarrow\; ___x = 3 \;\Longrightarrow\; x = 3/2
Why: Twenty five is 5 squared, so the exponent x becomes 2x when the base is rewritten. Equating with the 3 from 125 gives x equal to three halves. Checking: 25 to the power three halves is the cube of 5, which is 125.
Prediction
An exponential equation has bases that share no common power.
Predict first
What must be done instead?
Correct: Take a logarithm of both sides.
Why: The logarithm method works for every exponential equation, whether or not a common base exists, and it produces an exact answer expressible as a ratio of logarithms. The common-base shortcut is faster when available but is not always available, so the logarithm method is the general one.
Socratic
The step looks obvious and rests on a real property.
Discussion prompt
What property of exponential functions licenses equating the exponents?
Hint: Could two different exponents give the same output?
Answer:
An exponential function is one-to-one: no two different exponents give the same output. That is the horizontal line test from §1.1, which §4.2's graphs pass.
So if two powers of the same base are equal, their exponents must be equal — there is no other way to produce the same output.
The property is essential rather than decorative. For a function that were not one-to-one, such as squaring, equal outputs would not force equal inputs — which is exactly why solving a squared equation produces two answers and this one produces a single answer with no plus-or-minus.
Section
Section 2
Concept
When no common base is available, take a logarithm of both sides. The power property converts the unknown exponent into an ordinary coefficient, which can then be divided out.
Isolating first is not optional. Taking a logarithm of a sum containing an exponential achieves nothing, because there is no property for a logarithm of a sum — so the exponential must stand alone on its side before the logarithm is applied.
Figure (svg): The power property shown doing the essential work: an exponent containing the unknown brought down in front of a logarithm where it can be divided out
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 557-562
Picture it
Four lines, and the third is the one that solves the problem.
Figure (svg): The power property shown doing the essential work: an exponent containing the unknown brought down in front of a logarithm where it can be divided out
Once the exponent is in front, it is an ordinary coefficient and the equation is linear. Everything before that step is setup and everything after is arithmetic.
Worked example
Take a logarithm and bring the exponent down.
\[ \text{Solve } 5^x=30. \]
Note no common base exists
Why: Thirty is not a power of 5.
Take the natural logarithm of both sides
Why: Any base works; ln is convenient.
\[ \ln(5 ^{x}) = \ln 30 \]
Apply the power property
Why: Bring x down in front.
\[ x \ln 5 = \ln 30 \]
Divide
Why: By the coefficient of x.
\[ x = \ln 30 / \ln 5 \]
Figure (svg): The power property shown doing the essential work: an exponent containing the unknown brought down in front of a logarithm where it can be divided out
\[ x=\frac{\ln 30}{\ln 5}\approx 2.113 \]
Verify: check the size
Why: Five squared is 25 and 5 cubed is 125, so the answer should be between 2 and 3 and much closer to 2 — and 2.113 is. Note this is also the change-of-base formula's output for the logarithm base 5 of 30, which is exactly what the equation was asking for.
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 558-560
Ranking
For an exponential equation with no common base.
Put in order
Why: Isolating comes first because the logarithm can only be usefully applied to a lone exponential. Then the logarithm, then the power property that brings the exponent down, and finally the division that isolates the unknown as an ordinary coefficient.
Worked example
The exponential must stand alone first.
\[ \text{Solve } 3(2^x)+7=31. \]
Subtract the constant
Why: Getting the multiplied exponential alone.
\[ 3(2 ^{x}) = 24 \]
Divide by the coefficient
Why: Now the exponential is isolated.
\[ 2 ^{x} = 8 \]
Recognise a common base
Why: Eight is 2 cubed.
\[ 2 ^{x} = 2 ^{3} \]
Equate the exponents
Why: No logarithm needed after all.
\[ x = 3 \]
Figure (svg): The solution to Worked example isolate before taking the logarithm shown as a ladder of expressions, one row per legal move
\[ x=3 \]
Verify: check in the original
Why: Three times 2 cubed is 24, plus 7 is 31 — correct. Note that isolating first revealed a common base that was invisible in the original equation. Taking a logarithm of both sides at the start would have produced the logarithm of a sum, which cannot be expanded at all.
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 560-562
Error analysis
A student solves an exponential equation without isolating first.
Annotate
On: \( 2^x+5=13 \;\Longrightarrow\; \ln(2^x)+\ln 5=\ln 13 \)
Isolate the exponential before taking any logarithm. Applying a logarithm to a sum is the §4.5 error appearing in a new context, and it is why isolation is the first step rather than a tidying one.
Faded example
Solve 7 to the power x equals 100.
Fill in the blanks
x \ln 7 = \ln 100 \;\Longrightarrow\; x = \frac100}}7}}
Why: The power property brings x down in front of the logarithm of 7, and dividing isolates it. The result is about 2.37, which is sensible since 7 squared is 49 and 7 cubed is 343. The argument goes on top and the base underneath, as in the change-of-base formula.
Prediction
An exponential equation is solved twice, once using base ten logarithms and once using natural logarithms.
Predict first
Do the two give the same answer?
Correct: Yes, exactly the same.
Why: Both numerator and denominator scale by the same factor when the logarithm base changes, so the ratio is unchanged. This is the change-of-base formula again, and it means the choice of logarithm is purely a matter of convenience — whichever the calculator provides.
Explain it to yourself
Taking a logarithm of both sides only works on an isolated exponential.
Discussion prompt
Explain what goes wrong if the exponential is not isolated first.
Hint: What is on the side you are taking the logarithm of?
Answer:
If the exponential is added to something, the side is a sum. Taking a logarithm of a sum produces a logarithm that cannot be expanded, since §4.5 established there is no property for sums.
So the power property never becomes applicable, and the unknown stays stuck in the exponent inside an unexpandable logarithm. The move has made the equation worse rather than better.
Isolating first guarantees the logarithm is applied to a lone power, which is exactly the form the power property handles. Every step of the technique depends on that, which is why isolation is step one rather than a preliminary tidy.
Section
Section 3
Concept
When the unknown is inside a logarithm, condense the equation to a single logarithm and convert to exponential form, which frees the unknown.
The checking step is genuinely necessary here, not a formality. Condensing merges several logarithms into one whose domain is larger than the original's, so the converted equation can have solutions the original rejects — and only substitution finds them.
Figure (svg): Two solving strategies side by side: taking a logarithm of both sides when the unknown is in an exponent, and converting to exponential form when the unknown is inside a logarithm
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 562-566
Picture it
The right-hand column is this section, and it carries the checking warning.
Figure (svg): Two solving strategies side by side: taking a logarithm of both sides when the unknown is in an exponent, and converting to exponential form when the unknown is inside a logarithm
The exponential family needs no check because every real number is a legal exponent. The logarithmic family does, because arguments must be positive.
Worked example
Isolate, convert, solve, check.
\[ \text{Solve } \log_3(2x-1)=4. \]
Note the logarithm is already isolated
Why: Nothing added to it.
Convert to exponential form
Why: Base to the value equals the argument.
\[ 2 x - 1 = 3 ^{4} \]
Evaluate and solve
Why: Eighty one, then isolate.
\[ 2 x = 82, x = 41 \]
Check the argument is positive
Why: Twice 41 minus 1.
\[ 81 > 0,\text{ valid} \]
Figure (svg): The solution to Worked example one logarithm, converted shown as a ladder of expressions, one row per legal move
\[ x=41 \]
Verify: substitute into the original
Why: The argument is 81, and the logarithm base 3 of 81 is 4 — correct. The check also confirmed the argument was positive, which is the condition that could have invalidated the answer. Here it did not, but the check has to be run regardless.
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 563-564
Sorting
Every original argument must be positive at the candidate.
Sort into buckets
For the equation with log(x) plus log(x minus 4), sort each candidate.
Worked example
Two logarithms must become one before converting.
\[ \text{Solve } \log(x)+\log(x-3)=1. \]
Condense with the product property
Why: A sum of logarithms is a logarithm of a product.
\[ \log(x(x - 3)) = 1 \]
Convert to exponential form
Why: Base ten to the power 1.
\[ x ^{2} - 3 x = 10 \]
Solve the quadratic
Why: Rearrange and factor.
\[ (x - 5) (x + 2) = 0 \]
Read both candidates
Why: Five and negative 2.
\[ x = 5\text{ or } -2 \]
Figure (svg): A logarithmic equation whose algebra produces two candidate solutions, one of which makes an argument negative and must be discarded
\[ x=5 \]
Verify: check both in the ORIGINAL
Why: At 5: the arguments are 5 and 2, both positive, and the logarithms sum to the logarithm of 10, which is 1 — valid. At negative 2: the first argument is negative 2, which has no logarithm at all, so this candidate is discarded. The condensed equation accepted it because x times x minus 3 is positive there, but the original never did.
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 565-566
Trap
\[ \log x+\log(x-3)=1 \;\Longrightarrow\; x=5 \text{ or } x=-2, \text{ both reported} \]
Condense, convert and solve the quadratic
Why: The algebra is correct throughout and both roots are found.
Both values are given as solutions.
Only 5 works. At negative 2 the original equation asks for the logarithm of a negative number, which does not exist, so that candidate satisfies no equation at all.
It solves the condensed equation, whose single argument is a product that happens to be positive there. Condensing enlarged the domain and let it in.
Check every candidate in the original, not in the condensed version. This is the same structure as §3.8's extraneous solutions from squaring: a legitimate step that is not domain-preserving.
Faded example
Solve the equation setting the logarithm base 2 of the quantity x plus 5 equal to 3.
Fill in the blanks
x + 5 = 2^3} = 8 \;\Longrightarrow\; x = 3
Why: Converting to exponential form gives the argument equal to 2 cubed, which is 8, so x is 3. Checking: the argument is 8 and the logarithm base 2 of 8 is 3 — correct, and the argument is positive so the answer stands.
Prediction
Condensing two logarithms into one changes the expression's domain.
Predict first
In which direction does the domain change?
Correct: It gets larger, so extra solutions can appear.
Why: The original required each argument to be positive separately; the condensed version only requires their product to be positive, which also happens when both are negative. So the condensed equation accepts more inputs, and any solution in that extra region is extraneous.
Analogy
The same structure appeared in Chapter 3.
Match the pairs
Why: All three are legitimate algebraic steps that do not preserve the domain. Each can therefore produce candidates the original equation rejects, and in each case substituting into the original is the only remedy. Recognising the pattern makes the checking habit transfer between chapters.
Section
Section 4
Concept
If a single logarithm of the same base stands alone on each side, the arguments must be equal — because logarithms are one-to-one. The result still needs checking.
\[ \log_b M=\log_b N \;\Longrightarrow\; M=N \]
The property is the mirror of the one that let exponents be equated. Both functions are one-to-one, so both allow equal outputs to be traced back to equal inputs — which is why the two techniques in this chapter look so alike despite operating on opposite structures.
Figure (svg): Two solving strategies side by side: taking a logarithm of both sides when the unknown is in an exponent, and converting to exponential form when the unknown is inside a logarithm
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 564-567
Picture it
Both rest on the two functions being one-to-one.
Figure (svg): Two solving strategies side by side: taking a logarithm of both sides when the unknown is in an exponent, and converting to exponential form when the unknown is inside a logarithm
Equating exponents and equating arguments are the same move applied to the two inverse families, and both are licensed by the same property.
Worked example
One logarithm each side, same base.
\[ \text{Solve } \log_5(3x-2)=\log_5(x+8). \]
Check the bases match
Why: Both are base 5.
Equate the arguments
Why: Valid since logarithms are one-to-one.
\[ 3 x - 2 = x + 8 \]
Solve
Why: Collect and divide.
\[ 2 x = 10, x = 5 \]
Check both arguments
Why: Thirteen and thirteen.
Figure (svg): The solution to Worked example equate the arguments shown as a ladder of expressions, one row per legal move
\[ x=5 \]
Verify: confirm the arguments agree
Why: At x equal to 5 both arguments are 13, so both sides are the logarithm base 5 of 13 — equal, as required, and both positive so the solution stands. Note that the arguments coming out equal is exactly what equating them was asserting.
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 565-566
Sorting
One logarithm each side, same base.
Sort into buckets
Sort each equation.
Worked example
Each side must be a single logarithm before equating.
\[ \text{Solve } \log(x)+\log(2)=\log(x+3). \]
Condense the left side
Why: A sum becomes a product.
\[ \log(2 x) = \log(x + 3) \]
Equate the arguments
Why: Same base, one logarithm each side.
\[ 2 x = x + 3 \]
Solve
Why: Subtract x.
\[ x = 3 \]
Check the arguments
Why: Three, two and six, all positive.
Figure (svg): The solution to Worked example condense each side first shown as a ladder of expressions, one row per legal move
\[ x=3 \]
Verify: substitute into the original
Why: The left side is the logarithm of 3 plus the logarithm of 2, which is the logarithm of 6. The right is the logarithm of 6. They agree, and every argument is positive, so 3 is a genuine solution. Condensing before equating was essential — the arguments could not have been read off the unequal-length sides otherwise.
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 566-567
Error analysis
A student equates the arguments of two logarithms.
Annotate
On: \( \log_2(x)=\log_3(x+4) \;\Longrightarrow\; x=x+4 \)
Check the bases match before equating arguments. Different bases can produce equal values from different arguments, which is exactly why the property requires a shared base.
Faded example
Solve the equation with the natural logarithm of 4x on the left and of the quantity x plus 9 on the right.
Fill in the blanks
4x = x + 9 \;\Longrightarrow\; 3x = 9 \;\Longrightarrow\; x = 3
Why: Subtracting x from both sides gives 3x equal to 9, so x is 3. Checking: both arguments become 12, which is positive, so the solution is valid. Both sides being a single natural logarithm is what allowed the arguments to be equated.
Prediction
Equal logarithms force equal arguments.
Predict first
What property of the logarithm licenses that?
Correct: It is one-to-one, so no two arguments share a value.
Why: Being one-to-one is exactly the condition that different inputs give different outputs, so equal outputs must come from equal inputs. Being increasing implies it, but the property that does the work is the one-to-one condition itself — the same one that let exponents be equated earlier in this section.
Explain it
Equating exponents and equating arguments look like two rules.
Discussion prompt
Explain to a classmate why they are really the same idea.
Hint: What property does each rely on?
Answer:
Both rely on the function being one-to-one. If a function never gives the same output twice, then equal outputs must have come from equal inputs — which is precisely what both moves assert.
For the exponential the inputs are the exponents, so equal powers force equal exponents. For the logarithm the inputs are the arguments, so equal logarithms force equal arguments.
So there is one principle applied to two functions, and the two functions happen to be each other's inverse. That is why the chapter's two techniques mirror each other so closely: they are the same reasoning read from opposite ends.
Section
Section 5
Concept
The first decision in every problem is where the unknown sits. That determines which operation frees it, and whether a domain check will be needed.
The asymmetry in the last two points is the section's most useful practical fact. Exponential equations are safe to solve mechanically; logarithmic ones are not, and skipping the check there is a genuine source of wrong answers rather than merely poor practice.
Figure (svg): Two solving strategies side by side: taking a logarithm of both sides when the unknown is in an exponent, and converting to exponential form when the unknown is inside a logarithm
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 553-568
Picture it
Where the unknown sits decides the technique and the hazard.
Figure (svg): Two solving strategies side by side: taking a logarithm of both sides when the unknown is in an exponent, and converting to exponential form when the unknown is inside a logarithm
The red warning appears on only one side. That is the difference between the two families that matters most when working quickly.
Worked example
One glance decides the route.
\[ \text{Classify: } 4^{x+1}=9, \quad \log(x)+\log(x-1)=1, \quad 2^{3x}=16. \]
Look at the first
Why: Unknown in the exponent; 9 is not a power of 4.
Look at the second
Why: Unknown inside logarithms.
Look at the third
Why: Unknown in the exponent; 16 is a power of 2.
Note which needs checking
Why: Only the logarithmic one.
Figure (svg): Two solving strategies side by side: taking a logarithm of both sides when the unknown is in an exponent, and converting to exponential form when the unknown is inside a logarithm
\[ \text{log both sides; condense then convert (check!); common base} \]
Verify: confirm the third really has a common base
Why: Sixteen is 2 to the fourth, so the equation becomes 3x equal to 4 and x is four thirds. Spotting that in advance saved taking logarithms and produced an exact answer with no calculator, which is what makes the classification step worth doing before any algebra.
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 554-566
Sorting
Locate the unknown first.
Sort into buckets
Sort each equation.
Worked example
The exponential must be isolated even when a logarithm appears.
\[ \text{Solve } e^{2x}=7. \]
Note the base is e
Why: So the natural logarithm is the natural choice.
Take the natural logarithm of both sides
Why: The left collapses immediately.
\[ 2 x = \ln 7 \]
Note why it collapsed
Why: ln and e are inverses.
Divide
Why: By 2.
\[ x = \frac{\ln 7}{2} \]
Figure (svg): The solution to Worked example a mixed equation shown as a ladder of expressions, one row per legal move
\[ x=\tfrac{1}{2}\ln 7\approx 0.973 \]
Verify: check the size
Why: e is about 2.718, so e squared is about 7.39 — slightly more than 7, so the exponent 2x should be slightly less than 2 and x slightly less than 1. The answer 0.973 fits. Matching the logarithm's base to the exponential's is what made the left side collapse in one step.
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 560-562
Trap
\[ 2^x=5 \;\Longrightarrow\; x=\log_2 5\approx 2.32, \text{ then discarded for being irrational} \]
Solve and then apply a domain check
Why: The answer is checked against a domain requirement, by analogy with logarithmic equations.
The solution is rejected because it is not a whole number.
Every real number is a legal exponent, so there is no domain restriction to violate. An irrational answer is perfectly ordinary here.
Checking is harmless but never rejects anything in an exponential equation, because nothing about an exponent can be illegal.
The domain hazard belongs to the logarithmic family only. Knowing which family you are in tells you whether a check can change the answer, which saves both worry and time.
Prediction
An exponential equation with no common base has been solved.
Predict first
Can the domain check discard the answer?
Correct: No: every real number is a legal exponent.
Why: An exponential function's domain is unrestricted, so no candidate can violate it. Negative and irrational exponents are entirely ordinary. Only the logarithmic family carries a domain restriction, and that asymmetry is what makes the check compulsory there and optional here.
Faded example
Solve 4 to the power x equals 32, using the fastest route.
Fill in the blanks
2^5 = 2^5/2} \;\Longrightarrow\; 2x = ___ \;\Longrightarrow\; x = ___
Why: Both 4 and 32 are powers of 2, so rewriting over base 2 gives 2x equal to 5 and x equal to five halves. Spotting the common base avoided logarithms entirely and produced an exact answer. Checking: 4 to the power five halves is the fifth power of 2, which is 32.
Real world
Doubling times and half-lives are exactly these two families.
Discussion prompt
An investment grows exponentially and you want the time to double. Which family is that, and which is asking for the value after ten years?
Hint: In each, is time the unknown or is it given?
Answer:
The doubling question has time in the exponent and unknown, so it is an exponential equation solved by taking a logarithm. That is why doubling times are always logarithms of something.
The value-after-ten-years question gives the time, so it is an evaluation rather than an equation at all — substitute and compute.
This is §1.1's evaluating-against-solving distinction, and it explains why half-life formulas look so different from growth formulas even though they describe the same process. One is the model and the other is the model solved for its exponent, which is exactly what §4.7 makes routine.
Comparison
Fill the blanks from memory. The last row is the one that matters most in practice.
Comparison matrix
| unknown in an exponent | unknown inside a logarithm | |
|---|---|---|
| the move | take a logarithm of both sides | convert to exponential form |
| the property used | the power property | the definition |
| shortcut available | a common base, if both sides are powers of one number | equating arguments, if both sides are single logarithms |
| must isolate first | yes, the exponential | yes, condense to one logarithm |
| can produce extraneous answers | no | yes, always check |
The two families mirror each other in every row except the last, and that asymmetry comes entirely from the logarithm's restricted domain.
Pattern
Six steps, branching at the second.
Step 2's search for a common base is worth thirty seconds. When it succeeds the answer is exact and no calculator is needed, and when it fails nothing has been lost.
OpenStax Algebra and Trigonometry 2e, §6.6 Exponential and Logarithmic Equations §6.6
Check
Look for a common base first.
Check your understanding
What is the solution of 8 to the power x equals 32?
Answer: A
Why: Both sides are powers of 2: the equation becomes 2 to the 3x equals 2 to the fifth, so 3x is 5 and x is five thirds. Checking: 8 to the power five thirds is 2 to the fifth, which is 32.
Check
Isolate before taking a logarithm.
Check your understanding
For the equation 2 times 3 to the power x, plus 1, equals 19, what is the correct first step?
Answer: A
Why: The exponential must stand alone before a logarithm can usefully be applied, so the constant is removed first and then the coefficient. That leaves 3 to the x equal to 9, which has a common base and gives x equal to 2.
Check
Check against the original.
Check your understanding
Solving a logarithmic equation gives candidates 4 and negative 6, and the original contains the logarithm of x. What are the solutions?
Answer: A
Why: The logarithm of negative 6 does not exist, so that candidate makes the original equation meaningless and must be discarded. Only 4 survives, and it should be substituted to confirm it satisfies the equation.
Real world
Every half-life and doubling-time calculation is one of these equations.
Discussion prompt
Radiocarbon dating estimates an age from the fraction of carbon-14 remaining. Which family of equation is that, and why?
Hint: In the decay model, where does the age sit?
Answer:
The decay model puts the time in the exponent, and dating asks for the time given the remaining fraction. So the unknown is in the exponent and the equation is exponential.
Solving it means taking a logarithm of both sides so the power property brings the time down, then dividing — exactly this section's main technique.
And no domain check is needed, since any real number is a legal exponent. The only constraint is physical rather than mathematical: a negative age would be meaningless, and a fraction above 1 would indicate a measurement error rather than a mathematical impossibility.
Commit first
State your confidence along with your answer.
Predict first
Which family of equation can produce solutions that must be discarded?
Correct: Logarithmic, because arguments must be positive.
Why: Condensing merges logarithms into one whose domain is larger than the original's, so the converted equation can accept inputs the original rejects. Exponential equations have no such hazard, since every real number is a legal exponent — which is why the check is compulsory for one family and merely reassuring for the other.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why logarithmic equations need their answers checked and exponential ones do not.
Hint: What can be illegal in each case?
Answer:
A logarithm refuses negative and zero arguments, so a candidate that makes any original argument non-positive is not a solution at all — the equation is meaningless there.
Condensing hides that. Two logarithms merged into one have a single argument which can be positive even when both originals were negative, so the condensed equation accepts inputs the original never did.
An exponential refuses nothing: every real number is a legal exponent, so no candidate can be disqualified. A good explanation ties this back to §1.2's domain rules — the check exists because one family has a domain restriction and the other has none.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The fourth is the one that produces wrong answers rather than merely slow ones, since a missed check leaves an answer that is confidently and completely wrong. The second is the technique §4.7 will apply constantly, so fluency there pays off immediately.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw a decision flow starting from 'where is the unknown'. Branch to the exponential route and the logarithmic route, writing the steps of each. Mark clearly which branch requires checking and why. Then solve one equation of each kind, showing every step, and on the logarithmic one show a candidate being discarded.
If your flow marks the check on only one branch, with the reason written beside it, you have the asymmetry that this section exists to establish.
Recap
Five things, and the last one is the one that changes answers.
| if you remember one thing | it should be this |
|---|---|
| about choosing | locate the unknown; that decides the technique |
| about exponentials | isolate first, or the logarithm has a sum to deal with |
| about the power property | it is what brings the unknown exponent down |
| about logarithms | always check; condensing enlarges the domain |
Section 4.7 applies all of this to real models — growth, decay, half-life, Newton's law of cooling — where the equations of this section are what the questions come down to.
OpenStax, Precalculus, §4.6 Exponential and Logarithmic Equations §4.6, pp. 553-568 — everything on these slides traces back here
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