What rules can i use to simplify $b^\frac{\ln a}{\ln b}$ for $a,b>1$ ?
$\endgroup$4 Answers
$\begingroup$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 $\frac{\log a}{\log b} = \log_b a$.
$\endgroup$ $\begingroup$$b^{\frac{\ln a}{\ln b}}=b^{\log_ba}=a$.
$\endgroup$ 3 $\begingroup$$\dfrac{ln(a)}{ln(b)}$ = $\log_b a$
$b^{(\log_b a)} = a$
Hence the answer is $a$
$\endgroup$ $\begingroup$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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