# How can i simplify $b^\frac{\ln a}{\ln b}$?

What rules can i use to simplify $b^\frac{\ln a}{\ln b}$ for $a,b>1$ ?

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$$b^{\frac{\log a}{\log b}} = \exp\left(\log b \left(\frac{\log a}{\log b}\right)\right) = \exp( \log a ) = a$$

which may have the advantage of not requiring the knowledge that $\frac{\log a}{\log b} = \log_b a$.

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$b^{\frac{\ln a}{\ln b}}=b^{\log_ba}=a$.

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How do you get to $b^{log_{b}a}$ ? –  Devid Apr 3 '13 at 13:23
@Devid, let $\frac{\ln a}{\ln b}=t$, then $\ln a=\ln b^t$, and so $a=b^t$, yielding the result. –  Easy Apr 3 '13 at 13:26
@Devid. This technique involves a very useful result that isn't well enough known, namely that $(\log_pq)(\log_qr)=\log_pr$. To help remember this, there's a kind of cancellation involving the $q$s. A consequence is that log functions to different bases are essentially the same, up to constant multiples. –  Rick Decker Apr 3 '13 at 14:54

$\dfrac{ln(a)}{ln(b)}$ = $\log_b a$

$b^{(\log_b a)} = a$

Hence the answer is $a$

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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. $$x = a$$

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