Changing to a different exponent base raised to a logarithm

Question

I was playing around with logs, and I discovered the following, for $a,b,c,d > 0$

$$a^{\log(b^{{c}/{d}})}\ =\ b^{\frac{\log\left(a^{c}\right)}{d}}$$

Is this true for all cases? Why is this true? What is the general case for changing bases raised to a log?

Answer 1

Without $c,d$ :

$$X=e^{\log{(a)}\log{(b)}}=e^{\log{(b)}\log{(a)}}$$

$$X=(e^{\log{(a)}})^{\log{(b)}}=a^{\log{(b)}}$$
$$X=(e^{\log{(b)}})^{\log{(a)}}=b^{\log{(a)}}$$

Hence :
$$a^{\log{(b)}}=b^{\log{(a)}}$$

With $c,d$ :

$$Y=e^{\log{(a)}(c/d)\log{(b)}}=e^{\log{(b)}(c/d)\log{(a)}}$$

$$Y=(e^{\log{(a)}})^{(c/d)\log{(b)}}=a^{\log{(b^{c/d})}}$$
$$Y=(e^{\log{(b)}})^{(c/d)\log{(a)}}=b^{\log{(a^{c/d})}}$$

Hence :
$$a^{\log{(b^{c/d})}}=b^{\log{(a^{c/d})}}$$

Exponentiation & Logarithm :
https://en.wikipedia.org/wiki/Exponentiation
https://en.wikipedia.org/wiki/Exponentiation#Identities_and_properties
https://en.wikipedia.org/wiki/Logarithm
https://en.wikipedia.org/wiki/Logarithm#Logarithmic_identities
https://en.wikipedia.org/wiki/List_of_logarithmic_identities

Answer 2

$a^{\log(b^{{c}/{d}})} = b^{\frac{\log\left(a^{c}\right)}{d}}\iff$

$\log \left(a^{\log(b^{{c}/{d}})}\right) =\log \left( b^{\frac{\log\left(a^{c}\right)}{d}} \right)\iff$

$(\log a)\cdot \log(b^{{c}/{d}})=(\log b)\cdot \dfrac{\log(a^c)}{d}\iff$

$(\log a)\cdot (\log b)\cdot \dfrac c d =(\log b)\cdot \dfrac {c\cdot (\log a)}{d}\iff 1=1.$