How can i simplify $b^\frac{\ln a}{\ln b}$? What rules can i use to simplify $b^\frac{\ln a}{\ln b}$ for $a,b>1$ ?
 A: $\dfrac{ln(a)}{ln(b)}$ = $\log_b  a$
$b^{(\log_b  a)} = a$
Hence the answer is $a$
A: To get an equation that you can manipulate, let $x$ denote the expression you have.
$$x \; = \; b^{\frac{\ln a}{\ln b}}$$
Take the natural logarithm of both sides. Note that this is a natural thing to try, since doing this will convert the "harder" exponentiation operation on the right side into an "easier" multiplication operation.
$$\ln x \; = \; \ln \left( b^{\frac{\ln a}{\ln b}} \right)$$
Use a logarithm rule to rewrite the right side.
$$\ln x \; = \; \left( \frac{\ln a}{\ln b} \right) \cdot \ln b$$
Cancel common factors on the right side.
$$\ln x \; = \; \ln a$$
Use the fact that the natural logarithm function has the one-to-one property.
$$x = a$$
A: An alternative route to the answers already present is as follows:
$$b^{\frac{\log a}{\log b}} = \exp\left(\log b \left(\frac{\log a}{\log b}\right)\right) = \exp( \log a ) = a$$
which may have the advantage of not requiring the knowledge that $\frac{\log a}{\log b} = \log_b a$.
A: $b^{\frac{\ln a}{\ln b}}=b^{\log_ba}=a$.
