Exponential Equations

Solve exponential equations where an unknown exponent needs to be found using logarithms to transform them into solvable linear form.

ax=b    x=logaba^x = b \implies x = \log_a b

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What each symbol means

What Exponential Equations takes
aa
bb
xx
Exponential Equations
SymbolMeaning
aaThe base of the exponential expression; must be positive and not equal to 1, or the logarithm is undefined.
bbThe result value on the right side of the equation; must be positive, as you cannot take the logarithm of a negative number.
xxThe unknown exponent you are solving for; can be any real number, including negative or fractional values.

When to use it

Use exponential equations when you need to find an unknown exponent in a power equation.

Level

Usually taught in: Algebra II

Worked examples

1. Solve a simple exponential equation with integer base and result

Problem

Solve 2x=162^x = 16 for xx.
  1. 2x=162^x = 16

    This is the original exponential equation where the unknown exponent xx is what we need to find.

  2. log2(2x)=log2(16)\log_2(2^x) = \log_2(16)

    Take log2\log_2 (logarithm base 2) of both sides to bring down the exponent.

  3. x=log2(16)x = \log_2(16)

    Apply the property loga(ax)=x\log_a(a^x) = x, which says the base-aa logarithm of axa^x is just xx.

  4. x=4x = 4

    Evaluate log2(16)\log_2(16) by recognizing that 24=162^4 = 16, so log2(16)=4\log_2(16) = 4.

Answer: x=4x = 4

We converted the exponential equation into logarithmic form to isolate xx. Since 24=162^4 = 16, the answer checks: 24=162^4 = 16 ✓.

2. Solve an exponential equation with a fractional result

Problem

Solve 3x=193^x = \frac{1}{9} for xx.
  1. 3x=193^x = \frac{1}{9}

    This equation has a fractional result less than 1, which suggests xx will be negative since 3n=13n3^{-n} = \frac{1}{3^n}.

  2. log3(3x)=log3(19)\log_3(3^x) = \log_3\left(\frac{1}{9}\right)

    Take log3\log_3 of both sides to isolate the exponent.

  3. x=log3(19)x = \log_3\left(\frac{1}{9}\right)

    Apply loga(ax)=x\log_a(a^x) = x on the left side, leaving xx by itself.

  4. x=2x = -2

    Evaluate by recognizing that 32=132=193^{-2} = \frac{1}{3^2} = \frac{1}{9}, so log3(19)=2\log_3\left(\frac{1}{9}\right) = -2.

Answer: x=2x = -2

Fractional results in exponential equations lead to negative exponents. The negative answer makes sense: 32=193^{-2} = \frac{1}{9} ✓.

3. Solve a word problem with exponential growth

Problem

A basketball player's scoring per game increases exponentially according to the formula P=102nP = 10 \cdot 2^n, where PP is the points scored and nn is the game number. In which game will the player score 320 points?
  1. 102n=32010 \cdot 2^n = 320

    Set up the equation by substituting P=320P = 320 into the scoring formula.

  2. 2n=322^n = 32

    Divide both sides by 10 to isolate the exponential term.

  3. log2(2n)=log2(32)\log_2(2^n) = \log_2(32)

    Take log2\log_2 of both sides to bring down the exponent nn.

  4. n=log2(32)n = \log_2(32)

    Apply the property loga(ax)=x\log_a(a^x) = x to simplify the left side.

  5. n=5n = 5

    Evaluate log2(32)\log_2(32) by recognizing that 25=322^5 = 32.

Answer: n=5n = 5

The player will score 320 points in game 5. We can verify: P=1025=1032=320P = 10 \cdot 2^5 = 10 \cdot 32 = 320 ✓.

Common mistakes

Where Exponential Equations usually goes wrong
Answer came out wrong
Taking the logarithm of only one side, writing log(2x)=8\log(2^x) = 8 instead of log(2x)=log(8)\log(2^x) = \log(8).
Always take the logarithm of both sides: if ax=ba^x = b, write loga(ax)=loga(b)\log_a(a^x) = \log_a(b), then simplify.
Forgetting to apply the property loga(ax)=x\log_a(a^x) = x after taking the logarithm, leaving you with log3(8)\log_3(8) and not knowing what to do next.
After taking the logarithm of both sides, immediately simplify loga(ax)\log_a(a^x) to xx before attempting to evaluate the right side.
Using the wrong base for the logarithm, such as taking log10(3x)=log10(8)\log_{10}(3^x) = \log_{10}(8) instead of log3(3x)=log3(8)\log_3(3^x) = \log_3(8).
Match the logarithm base to the exponential base: for ax=ba^x = b, use loga\log_a on both sides, not a different base.
The mistakeWhy it is wrongThe fix
Taking the logarithm of only one side, writing log(2x)=8\log(2^x) = 8 instead of log(2x)=log(8)\log(2^x) = \log(8).Logarithm is a function, and to preserve equality you must apply it to both sides of the equation, just like with any algebraic operation.Always take the logarithm of both sides: if ax=ba^x = b, write loga(ax)=loga(b)\log_a(a^x) = \log_a(b), then simplify.
Forgetting to apply the property loga(ax)=x\log_a(a^x) = x after taking the logarithm, leaving you with log3(8)\log_3(8) and not knowing what to do next.The logarithm property is what actually isolates xx and makes the problem solvable; skipping it leaves xx trapped inside a logarithm.After taking the logarithm of both sides, immediately simplify loga(ax)\log_a(a^x) to xx before attempting to evaluate the right side.
Using the wrong base for the logarithm, such as taking log10(3x)=log10(8)\log_{10}(3^x) = \log_{10}(8) instead of log3(3x)=log3(8)\log_3(3^x) = \log_3(8).The logarithm property loga(ax)=x\log_a(a^x) = x only works when the base of the logarithm matches the base of the exponential.Match the logarithm base to the exponential base: for ax=ba^x = b, use loga\log_a on both sides, not a different base.

Tips and when to use something else

  • If both sides of the equation already have the same base, such as 2x=252^x = 2^5, you can skip logarithms and equate the exponents directly to get x=5x = 5.
  • For equations like 2x=82^x = 8, try to rewrite the right side as a power of the left base (for example, 8=238 = 2^3); this makes solving possible without a calculator.
  • When bases do not match and you need an exact answer, the solution is a logarithm; use Logarithmic Equations if your problem asks you to simplify x=loga(b)x = \log_a(b) further or evaluate it numerically.
  • Exponential equations model real-world situations like population growth, radioactive decay, bacterial growth, and compound interest; recognizing when a word problem leads to an exponential equation is key to setting it up correctly.

Frequently asked questions

How do I know which base to use for the logarithm?
Use the same base as the base of the exponential term. If your equation is 2x=502^x = 50, take log2\log_2 of both sides. The base of the logarithm must match the base of the exponential so that the property loga(ax)=x\log_a(a^x) = x applies correctly.
Can I have a negative or zero base in an exponential equation?
No. The base aa must be positive and not equal to 1; otherwise the logarithm is undefined. Also, bb must be positive because you cannot take the logarithm of a negative number in real-number arithmetic.
What if I can't evaluate the logarithm by hand, like in 2x=502^x = 50?
Use the change of base formula to convert to a common or natural logarithm: x=log2(50)=log(50)log(2)5.64x = \log_2(50) = \frac{\log(50)}{\log(2)} \approx 5.64. This allows you to use a calculator to find a decimal approximation of the answer.
What's the difference between an exponential equation and a logarithmic equation?
An exponential equation has the unknown in the exponent, like 2x=82^x = 8. A logarithmic equation has the unknown inside a logarithm, like log2(x)=3\log_2(x) = 3. They are inverses: solving 2x=82^x = 8 with logarithms gives x=log2(8)x = \log_2(8), which is a logarithmic statement.

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Reviewed 2026-09-18