Interesting Mathematical Problems

Powers with logarithmic exponents

An expression where a logarithm sits in the exponent. The solution uses the power rule for logarithms and the identity a to the power of log base a of b equals b.

A logarithm in the exponent looks alarming, but it is the shortest road to the answer. There is one key: a raised to the power of log base a of b is simply b. All you have to do is bring the base of the power and the base of the logarithm to the same number.

Calculate

3^{\log_{9} 27} + \log_{\tfrac{1}{3}} 9 - \left(\sqrt{3}\right)^{\log_{3} 16}

Solution

3^{\tfrac{3}{2}} + (-2) - \left(\sqrt{3}\right)^{\log_{3} 16}

\left(3^{\tfrac{1}{2}}\right)^{3} - 2 - 4

\left(\sqrt{3}\right)^{3} - 6 = 3\sqrt{3} - 6

Notes in the margin

\log_{9} 27 = \log_{9} 3^{3} = 3 \log_{9} 3

\log_{a} x^{p} = p \cdot \log_{a} x

\log_{9} 3 = \tfrac{1}{2} \;\Rightarrow\; 9^{\tfrac{1}{2}} = 3

3 \log_{9} 3 = 3 \cdot \tfrac{1}{2} = \tfrac{3}{2}

\log_{\tfrac{1}{3}} 9 = -2 \;\Rightarrow\; \left(\tfrac{1}{3}\right)^{-2} = 9

\sqrt{3} = 3^{\tfrac{1}{2}} \;\Rightarrow\; \left(3^{\tfrac{1}{2}}\right)^{\log_{3} 16} = 3^{\tfrac{1}{2}\log_{3} 16} = 3^{\log_{3} 16^{\tfrac{1}{2}}} = 3^{\log_{3} 4} = 4

a^{\log_{a} b} = b

Frequently asked questions

Why does a to the power of log base a of b equal b?

Because the logarithm base a of b is by definition the exponent you must raise a to in order to get b. Raising a to that exponent therefore has to give b back.

What is the logarithm base one third of nine?

Minus two, because one third to the power of minus two is nine. A base smaller than one makes the logarithm of a number greater than one negative.

How do you simplify the logarithm of a power?

The exponent comes out in front: the logarithm of x to the power p equals p times the logarithm of x. That is why log₉27 is written as log₉3³, which is 3 log₉3.

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