9

I have tried using the change of base formula, but can't quite complete the equality:

$$ a^{\log{b}} \\ a^{\frac{\log_a{b}}{\log_a{a}}} $$

How do I get the base of the exponent to be b?

David Faux
  • 3,425
  • Hello David: I really thank you for posting this problem to get great answers, and I am up-voting accordingly! The result of this question was used (and linked) in the linked Stack Exchange Math Question, "Algebraically Simplify $n =\left( \frac{ \left(5\right)^{\log 5}} { \left(2 \right)^{\log 2}} \right) ^{ \frac{1} {\log 2-\log 5 } } = \frac{1}{10} $" Once again, thank you for your usesful question and all the useful answers from below! – Stephen Elliott Mar 07 '24 at 10:02

3 Answers3

17

Since $x = e^{\log x}$

\begin{align*} a^{\log b} = \left(e^{\log a}\right) ^{\log b} = e^{\log a \log b} = e^{\log b \log a} = \left(e^{\log b}\right)^{\log a} = b^{\log a} \end{align*}

5

Let $x=a^{\log b}$ and $y=b^{\log a}$. Then $\log x= \log a$. $\log b= \log y$. So, $x=y$.

voldemort
  • 13,182
2

Let (capital) $B$ be the base of the logarithms. Then $$ a^{\log_B b} = \Big(B^{\log_B a}\Big)^{\log_B b} = B^{(\log_B a)(\log_B b)}, $$ and that is clearly symmetric in $a$ and $b$. Or you could just go on from there: $$ \cdots= \Big(B^{\log_B b}\Big)^{\log_B a} = b^{\log_B a}. $$

Michael Hardy
  • 1