I learned all the scripts provided, yet I simply cannot find a way to simplify the following equations:

$$4^{log_{2}9}$$ What I've thought so far: 9 may be written as $3^2$, so maybe we could do something with the base $log_2$, but therefore it would have to be $2^3$, not the other way round - cannot think of any other approach.

EDIT: See How $a^{\log_b x} = x^{\log_b a}$? top answer (basically it explains how to transform $a^{\log_b(x)}$ to $x^{\log_b(a)}$)

$$8^{\frac{1}{3}log_{6}4}$$ What I've done so far: $8^{1/3}=2 \Rightarrow 8^{\frac{1}{3}log_{6}4} = 2^{log_{6}4}$ - but I have no idea how to simplify this term. (Maybe something with $4=2^2$ and $6=3*2$ and do something with the $2$s, but I don't know of any rule that allows you to "split" bases)

EDIT: turned out to be a typo, it should have read like this $8^{\frac{1}{3}log_{2}4}$ which basically is $2^2 = 4$

$$e^{\frac{1}{2}log9}$$ No idea what to do here. (Maybe try to "bring down" the natural logarithm, but how?)

EDIT: \begin{align*} e^{\frac{1}{2}log9} &= 9^{\frac{1}{2}log9\ log_{9}e} \\\ &= 9^{\frac{1}{2}\frac{log_{9}9}{log_{9}e}log_{9}e} \\\ &= 3 \end{align*} In words: change base, then change base in the exponent and cancel (can you put it like that in English?) so that only 3 remains.

$$b^{4log_{b}x}$$ Same like above - try to bring down the logarithm with the base $b$.

EDIT: (In thoughts) isolate the $log$, so that $b^{log_{b}x}$ remains $\Rightarrow$ this is $x$ $\Rightarrow$ the final result should be $x^4.

$$e^{3log\ 8x}$$ Same here with natural logarithm again.

EDIT: Same like the previous example, final result should be $(8x)^3$.

To prevent any misunderstandings: I'm not asking for solutions or step-by-step-instructions, just want to know if I'm heading in the right direction and if there's any rule/approach I overlooked which can be applied here. (Sadly we didn't do much logarithm-related stuff in school and at university it's considered to be foreknowledge, so they don't explain it any more)


2 Answers 2


A start: $4=2^2$ and therefore $4^x=(2^2)^x=2^{2x}$.

  • $\begingroup$ Thanks for the quick answer :) Any way to further simplify $2log_{2}3^2$? $\endgroup$
    – Peter
    Oct 15, 2012 at 19:25
  • 1
    $\begingroup$ Yes, two ways. $2\log_a x=\log_a x^2$. Or better (I think) $2^{2\log_2 9}=2^{(\log_2 9)(2)}=(2^{\log_2 9})^2$. We are using $a^{bc}=(a^b)^c$. $\endgroup$ Oct 15, 2012 at 19:28
  • $\begingroup$ Ahh, god I should have thought of that... Thanks again :) $\endgroup$
    – Peter
    Oct 15, 2012 at 19:30

There is a simpler way

\begin{align}\large n^{\log_bx} = x^{\log_bn}\end{align} Now we have for this question, \begin{align}\large 4^{\log_29} = 9^{\log_24}\end{align} As the OP is not asking for solution, please proceed from here.


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