Power Rule for Logarithms

The Power Rule for Logarithms lets you pull an exponent out of a logarithm as a multiplier, making complex logs easier to solve and simplify.

loga(xn)=nlogax\log_a(x^n) = n \log_a x

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What each symbol means

What Power Rule for Logarithms takes
aa
xx
nn
Power Rule for Logarithms
SymbolMeaning
aaThe base of the logarithm; must be positive and not equal to 1, or the logarithm is undefined.
xxThe argument of the logarithm; must be positive or the logarithm has no real value.
nnThe exponent on the argument; can be any real number (positive, negative, or fractional), and determines how many times the logarithm is multiplied by.

When to use it

Use this when you see an exponent inside a logarithm that you need to simplify or extract.

Level

Usually taught in: Algebra II

Worked examples

1. Simplify a logarithm with a small integer exponent

Problem

Simplify log2(83)\log_2(8^3)
  1. log2(83)=3log2(8)\log_2(8^3) = 3\log_2(8)

    Apply the Power Rule for Logarithms to pull the exponent out to the front as a multiplier.

  2. 3log2(8)=333\log_2(8) = 3 \cdot 3

    Evaluate log2(8)\log_2(8) by asking what power of 2 gives 8; since 23=82^3 = 8, we have log2(8)=3\log_2(8) = 3.

  3. 33=93 \cdot 3 = 9

    Multiply to get the final answer.

Answer: 99

We used the Power Rule to move the exponent outside the logarithm, then evaluated the resulting logarithm by recognizing that 8=238 = 2^3.

2. Simplify a logarithm with a negative fractional exponent

Problem

Simplify log4(161/2)\log_4(16^{-1/2})
  1. log4(161/2)=12log4(16)\log_4(16^{-1/2}) = -\frac{1}{2}\log_4(16)

    Apply the Power Rule for Logarithms to pull out the exponent, which is negative and fractional.

  2. 12log4(16)=122-\frac{1}{2}\log_4(16) = -\frac{1}{2} \cdot 2

    Evaluate log4(16)\log_4(16) by asking what power of 4 gives 16; since 42=164^2 = 16, we have log4(16)=2\log_4(16) = 2.

  3. 122=1-\frac{1}{2} \cdot 2 = -1

    Multiply to simplify.

Answer: 1-1

This problem combines the Power Rule with a negative fractional exponent, showing that the rule works for any real exponent, not just positive integers.

3. Simplify logarithmic equations with two ticket tiers

Problem

A concert venue analyzes ticket sales using the formula log3(Q2)=log3(D2)\log_3(Q^2) = \log_3(D^2), where QQ is the quantity of VIP tickets and DD is the quantity of discount tickets. Simplify both sides to find the relationship between QQ and DD.
  1. log3(Q2)=2log3(Q)\log_3(Q^2) = 2\log_3(Q)

    Apply the Power Rule to the left side, pulling the exponent 2 out to become a coefficient.

  2. log3(D2)=2log3(D)\log_3(D^2) = 2\log_3(D)

    Apply the Power Rule to the right side in the same way, pulling out the exponent from the discount tier.

  3. 2log3(Q)=2log3(D)log3(Q)=log3(D)2\log_3(Q) = 2\log_3(D) \Rightarrow \log_3(Q) = \log_3(D)

    Divide both sides by 2 to isolate the simplified logarithms.

Answer: log3(Q)=log3(D)\log_3(Q) = \log_3(D)

By applying the Power Rule to both sides, we simplified the equation and discovered that the quantities of VIP and discount tickets must be equal since their logarithms are equal.

Common mistakes

Where Power Rule for Logarithms usually goes wrong
Answer came out wrong
loga(xn)=(logax)n\log_a(x^n) = (\log_a x)^n
The correct form is loga(xn)=nlogax\log_a(x^n) = n \log_a x (multiply nn by the logarithm).
loga(xn)=logan(x)\log_a(x^n) = \log_{a^n}(x)
The exponent stays outside as a coefficient: loga(xn)=nlogax\log_a(x^n) = n\log_a x.
loga(xn)=loga(x+n)\log_a(x^n) = \log_a(x + n)
The correct form is loga(xn)=nlogax\log_a(x^n) = n\log_a x (multiply by the exponent, don't add it).
The mistakeWhy it is wrongThe fix
loga(xn)=(logax)n\log_a(x^n) = (\log_a x)^nThis reverses the rule; you multiply by the exponent, not raise to it.The correct form is loga(xn)=nlogax\log_a(x^n) = n \log_a x (multiply nn by the logarithm).
loga(xn)=logan(x)\log_a(x^n) = \log_{a^n}(x)The Power Rule only applies to the argument (inside), never to the base; changing the base changes the entire logarithm's meaning.The exponent stays outside as a coefficient: loga(xn)=nlogax\log_a(x^n) = n\log_a x.
loga(xn)=loga(x+n)\log_a(x^n) = \log_a(x + n)Exponents don't become addition when you leave the logarithm; the Power Rule specifically says to pull the exponent out and multiply.The correct form is loga(xn)=nlogax\log_a(x^n) = n\log_a x (multiply by the exponent, don't add it).

Tips and when to use something else

  • The Power Rule works for any real exponent—positive, negative, fractional—as long as x>0x > 0.
  • Don't confuse this with the Product Rule for Logarithms (loga(xy)=loga(x)+loga(y)\log_a(xy) = \log_a(x) + \log_a(y)), which applies when you have multiplication inside the log, not an exponent.
  • Use this rule in reverse too: if you see nloga(x)n\log_a(x), you can write it as loga(xn)\log_a(x^n) to combine logarithms.
  • When solving logarithmic equations, the Power Rule often appears before you reach for the Change of Base Formula or work with natural logarithms.

Frequently asked questions

Can I use the Power Rule if the exponent is negative or fractional?
Yes. The Power Rule applies to any real exponent—negative, fractional, or irrational. For example, log2(x3)=3log2(x)\log_2(x^{-3}) = -3\log_2(x) and log5(x1/2)=12log5(x)\log_5(x^{1/2}) = \frac{1}{2}\log_5(x). The only requirement is that the argument xx must be positive.
How is the Power Rule different from the Product Rule for Logarithms?
The Power Rule applies when you have an exponent inside the logarithm, like loga(xn)\log_a(x^n); it lets you pull that exponent out as a multiplier. The Product Rule applies when you multiply two things inside the logarithm, like loga(xy)\log_a(xy); it lets you split that into a sum. They're both simplification tools but handle different situations.
Can I apply the Power Rule to the base of the logarithm?
No. The Power Rule only applies to the argument (the thing inside the logarithm), not to the base. If you tried to rewrite loga2(x)\log_{a^2}(x) using the Power Rule, you would break the rule and get the wrong answer. Always check that the exponent is inside the logarithm before using this rule.
Why do I need the Power Rule if I can just calculate the exponent and then take the logarithm?
In algebra, you usually work with variables and expressions, not numbers you can evaluate directly. The Power Rule lets you simplify expressions like log2(x5)\log_2(x^5) into 5log2(x)5\log_2(x) so you can solve equations, combine logarithms, or integrate in calculus later. It's a tool for algebraic manipulation, not just computation.

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Reviewed 2026-09-18