Power Rule for Logarithms

Learn how the power rule moves an exponent in front of a logarithm, how roots become fractional powers, and how to expand and condense logs with all three rules.

Advanced25 min18 problems
Practice Problems

Definition

The power rule for logarithms says that an exponent inside a logarithm can be moved to the front as a factor:
log⁡b(Mp)=p⋅log⁡bM\log_b(M^p) = p \cdot \log_b M
It works for any real exponent pp, as long as M>0M > 0, b>0b > 0 and b≠1b \neq 1.
Why it is true: let x=log⁡bMx = \log_b M, so M=bxM = b^x. Then Mp=(bx)p=bpxM^p = (b^x)^p = b^{px}, and the logarithm of bpxb^{px} is simply pxpx. So log⁡b(Mp)=p⋅x=p⋅log⁡bM\log_b(M^p) = p \cdot x = p \cdot \log_b M.
Roots are powers too. Since Mn=M1/n\sqrt[n]{M} = M^{1/n}:
log⁡bMn=1nlog⁡bM\log_b \sqrt[n]{M} = \frac{1}{n} \log_b M
Example: log⁡2(85)=5⋅log⁡28=5⋅3=15\log_2(8^5) = 5 \cdot \log_2 8 = 5 \cdot 3 = 15, and log⁡1000=12⋅3=32\log \sqrt{1000} = \frac{1}{2} \cdot 3 = \frac{3}{2}.
Together with the product rule (log⁡b(MN)=log⁡bM+log⁡bN\log_b(MN) = \log_b M + \log_b N) and the quotient rule (log⁡bMN=log⁡bM−log⁡bN\log_b \frac{M}{N} = \log_b M - \log_b N), the power rule lets you expand one logarithm into many simple ones, or condense several logarithms into one.

Try it now

Which expression is equal to log⁡3(x5)\log_3(x^5) for x>0x > 0?

Worked Examples

Evaluate log⁡2(85)\log_2(8^5) without a calculator.
1

Move the exponent to the front

log⁡2(85)=5⋅log⁡28\log_2(8^5) = 5 \cdot \log_2 8 → 5log⁡285 \log_2 8
2

Evaluate the simpler logarithm

23=82^3 = 8, so log⁡28=3\log_2 8 = 3
3

Multiply

5⋅3=155 \cdot 3 = 15

Common Mistakes

Treating (log⁡M)p(\log M)^p as plog⁡Mp \log M

Why it's wrong: The power rule only works when the exponent is on the argument, inside the logarithm. In (log⁡M)p(\log M)^p the whole logarithm is raised to a power. Example: (log⁡100)3=23=8(\log 100)^3 = 2^3 = 8, but 3log⁡100=63 \log 100 = 6.

Correct: log⁡(Mp)=plog⁡M\log(M^p) = p \log M, while (log⁡M)p(\log M)^p cannot be simplified with the power rule. Look where the exponent sits before you move it.

Applying the exponent to the whole argument: log⁡(3x2)=2log⁡(3x)\log(3x^2) = 2 \log(3x)

Why it's wrong: In 3x23x^2 only xx is squared, not 33. Check with x=10x = 10: log⁡300≈2.48\log 300 \approx 2.48, but 2log⁡30≈2.952 \log 30 \approx 2.95.

Correct: Split the product first: log⁡(3x2)=log⁡3+log⁡x2=log⁡3+2log⁡x\log(3x^2) = \log 3 + \log x^2 = \log 3 + 2 \log x.

Turning a square root into a factor of 2: log⁡x=2log⁡x\log \sqrt{x} = 2 \log x

Why it's wrong: A square root is the power 12\frac{1}{2}, not the power 22. The index of the root goes into the denominator.

Correct: log⁡x=log⁡x1/2=12log⁡x\log \sqrt{x} = \log x^{1/2} = \frac{1}{2} \log x, and log⁡x3=13log⁡x\log \sqrt[3]{x} = \frac{1}{3} \log x.

Condensing 3log⁡x−log⁡y3 \log x - \log y into log⁡3xy\log \frac{3x}{y}

Why it's wrong: A coefficient in front of a logarithm is an exponent of the argument, not a factor of it.

Correct: First move the coefficient up: 3log⁡x=log⁡x33 \log x = \log x^3. Then 3log⁡x−log⁡y=log⁡x3y3 \log x - \log y = \log \frac{x^3}{y}.

Interactive Visual

Function graph

-10-50510-4-2024xyx = 0

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Asymptote

Table of values

x-100.11101000
log(x²)2-2026
2 log(x)undef.-2026
(log x)²undef.1019

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Practice Problems

18 problems
Problem 1 of 18
Easy
Which expression is equal to log⁡3(x5)\log_3(x^5) for x>0x > 0?

Why It Matters

The power rule is the property that makes logarithms truly useful:
  • Solving exponential equations: in 1.05t=21.05^t = 2 the unknown sits in the exponent. Taking the logarithm and using the power rule gives t⋅log⁡1.05=log⁡2t \cdot \log 1.05 = \log 2, so the exponent comes down and becomes a normal factor.
  • Huge numbers: log⁡(264)=64log⁡2≈19.3\log(2^{64}) = 64 \log 2 \approx 19.3, which tells you at once that 2642^{64} has 20 digits, without computing it.
  • Science scales: decibels, pH and the Richter scale all use logarithms; the power rule explains why squaring an intensity ratio doubles the change in decibels.
  • Calculus: before you differentiate a complicated product of powers, you expand its logarithm with the power rule.
It is the last of the three main log rules, and you will use all three together in almost every logarithm problem from now on.

Real World Applications

Loudness in Decibels

The change in sound level is $\Delta L = 10 \log \frac{I_2}{I_1}$ decibels, where $\frac{I_2}{I_1}$ is how many times the intensity grows. The power rule turns powers of 10 into simple multiplication.

Example:

Doubling your distance from a speaker divides the intensity by 22=42^2 = 4, so ΔL=10log⁡(2−2)=−20log⁡2≈−6\Delta L = 10 \log(2^{-2}) = -20 \log 2 \approx -6 dB.

1Try It Yourself

You stand 2 m from a loudspeaker and walk back to 20 m. Sound intensity falls with the square of the distance, so it is divided by (202)2\left(\frac{20}{2}\right)^2. The level drops by 10log⁡((202)2)10 \log\left(\left(\frac{20}{2}\right)^2\right) decibels.

By how many decibels does the sound level drop?

Step 1: Write the mathematical expression

Compute 10log⁡(102)10 \log(10^2):

Counting Digits of Huge Numbers

A whole number $N$ has $\lfloor \log N \rfloor + 1$ digits. With the power rule you can count the digits of enormous powers without computing them.

Example:

log⁡(210)=10log⁡2≈3.01\log(2^{10}) = 10 \log 2 \approx 3.01, so 210=10242^{10} = 1024 has 3+1=43 + 1 = 4 digits.

2Try It Yourself

Computers often use 6464-bit numbers, and 2642^{64} is the number of different values they can store. Use log⁡2≈0.301\log 2 \approx 0.301.

How many digits does 2642^{64} have?

Step 1: Write the mathematical expression

Number of digits: ⌊64log⁡2⌋+1\lfloor 64 \log 2 \rfloor + 1

Population Growth

A population that grows by a fixed percentage every year is an exponential function of time. To find when it reaches a target, the power rule brings the time down from the exponent.

Example:

At 10% growth per year, a population doubles when 1.1t=21.1^t = 2: tlog⁡1.1=log⁡2t \log 1.1 = \log 2, so t=log⁡2log⁡1.1≈7.3t = \frac{\log 2}{\log 1.1} \approx 7.3 years.

3Try It Yourself

A town has 20000 inhabitants and grows by 3% per year. After tt years it has 20000⋅1.03t20000 \cdot 1.03^t inhabitants.

After how many full years does the town have at least 30000 inhabitants?

Step 1: Write the mathematical expression

Round t=log⁡1.5log⁡1.03t = \frac{\log 1.5}{\log 1.03} up to a whole year:

Key Takeaways

  • 1Power rule: log⁡b(Mp)=plog⁡bM\log_b(M^p) = p \log_b M for M>0M > 0 and any real exponent pp
  • 2Roots are fractional powers: log⁡bMn=1nlog⁡bM\log_b \sqrt[n]{M} = \frac{1}{n} \log_b M
  • 3A reciprocal is the power −1-1: log⁡b1M=−log⁡bM\log_b \frac{1}{M} = -\log_b M
  • 4Expanding: split products and quotients first, then bring every exponent to the front
  • 5Condensing: turn every coefficient into an exponent first, then combine sums into products and differences into quotients
  • 6(log⁡M)p(\log M)^p is not plog⁡Mp \log M: the exponent must be inside the logarithm

Frequently Asked Questions

A logarithm is an exponent. If log⁡bM=x\log_b M = x, then M=bxM = b^x and Mp=bpxM^p = b^{px}. Raising to the power pp multiplies the exponent by pp, so the logarithm (the exponent) is multiplied by pp too.
A logarithm is an exponent. If log⁡bM=x\log_b M = x, then M=bxM = b^x and Mp=bpxM^p = b^{px}. Raising to the power pp multiplies the exponent by pp, so the logarithm (the exponent) is multiplied by pp too.
No. The power rule works for any real exponent: negative (log⁡b1M2=−2log⁡bM\log_b \frac{1}{M^2} = -2 \log_b M), fractional (log⁡bM3=13log⁡bM\log_b \sqrt[3]{M} = \frac{1}{3} \log_b M) and even irrational ones.
Only for x>0x > 0. The function log⁡(x2)\log(x^2) also exists for negative xx (for example log⁡((−10)2)=log⁡100=2\log((-10)^2) = \log 100 = 2), but 2log⁡x2 \log x does not. For every x≠0x \neq 0 the exact statement is log⁡(x2)=2log⁡∣x∣\log(x^2) = 2 \log |x|. The graph above shows both curves.
Expand when you want simple pieces, for example to approximate log⁡72\log 72 from log⁡2\log 2 and log⁡3\log 3. Condense when you want one logarithm, for example to solve an equation like 2log⁡x−log⁡4=22 \log x - \log 4 = 2 (see Solving Logarithmic Equations).

Glossary

Power rule
The logarithm property log⁡b(Mp)=plog⁡bM\log_b(M^p) = p \log_b M
Exponent
The number that tells how many times the base is used as a factor, as in MpM^p
Fractional exponent
An exponent written as a fraction; M1/nM^{1/n} is the nn-th root of MM
Expand
Rewrite one logarithm as a sum and difference of simpler logarithms
Condense
Rewrite several logarithms as a single logarithm
Coefficient
A number multiplying a logarithm, as the 33 in 3log⁡x3 \log x; the power rule turns it into an exponent

Formula Card

Power rule

log⁡b(Mp)=plog⁡bM\log_b(M^p) = p \log_b M

The exponent of the argument moves to the front

Root rule

log⁡bMn=1nlog⁡bM\log_b \sqrt[n]{M} = \frac{1}{n} \log_b M

Because $\sqrt[n]{M} = M^{1/n}$

Reciprocal

log⁡b1M=−log⁡bM\log_b \frac{1}{M} = -\log_b M

Because $\frac{1}{M} = M^{-1}$

Product rule

log⁡b(MN)=log⁡bM+log⁡bN\log_b(MN) = \log_b M + \log_b N

Reminder: products become sums

Quotient rule

log⁡bMN=log⁡bM−log⁡bN\log_b \frac{M}{N} = \log_b M - \log_b N

Reminder: quotients become differences

Today's challengeOne problem, three tries

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