Logs with exponential bases I know that $e^{\log_{e^2} (16)}$ is 4, but I can only get this far:
$$e^{\log_{e^2}(4^2)}$$
$$e^{2\log_{e^2}(4)}$$
I need some way to cancel the 2's so I can get
$$e^{\log_e(4)}$$
but I don't the identity or rule that cancels the 2's.
 A: $$\exp{\log_{e^2}(16)} = \exp\frac{\log{16}}{\log(e^2)} = \exp\frac{\log{16}}{2} = \exp{\log(16^{1/2})} = 16^{1/2} = 4.$$
I used the following rules: $\log_a(x) = \frac{\log(x)}{\log(a)}$, and $c\log(x) = \log(x^c)$.
A: To answer your question about "cancelling". 
$b^{n *\log_{b^n} x} = (b^n)^{\log_{(b^n)} x} = x$ 
====
$e^{\log_{e^2}16}= e^{\log_{e^2}4^2}= e^{2*{\log_{e^2}4}}=({e^2})^{\log_{(e^2)}4}=4$
Or better yet:
$e^{\log_{e^2}16} =e^{\frac {\ln 16}{\ln e^2}} = e^{\frac {\ln 2^4}{2}}=e^{\frac {4\ln 2}{2}} = e^{2\ln 2} = (e^{\ln 2})^2 =2^2 = 4$
Or....$e^{\log_{e^2}16} =e^{\frac {\ln 16}{\ln e^2}} = e^{\frac {\ln 4^2}{2}}=e^{\frac {2\ln 4}{2}} = e^{\ln 4} = 4$.
Etc. etc. etc. etc.
A: Let's reason out what the expression for the exponent represents.  That is, denote
$$
q = \log_{e^2} 16
$$
This is equivalent to
$$
(e^2)^q = 16
$$
Since, in general for positive $a, b, c, (a^b)^c = a^{bc} = (a^c)^b$, this becomes
$$
(e^q)^2 = 16
$$
or finally
$$
e^q = 4
$$
where we can eliminate $e^q = -4$ because for real $q$, $e^q$ is necessarily positive.  And we're done!
A: Instead of thinking about "cancelling the 2s", try it this way:  You can write $e^{2\log_{e^2}(4)}$ as $(e^2)^{log_{e^2}(4)}$.  Now use the fact that for any base $b$, $b^{\log_b(x)}=x$.
A: You can use the following formula for logarithms $$\log_{a^q}b=\tfrac1q \log_ab$$
A: Suppose $a^x=y$; then
$$
x\log a=\log y
$$
(the unadorned $\log$ denotes the natural logarithm). This means that
$$
\log_a x=\frac{\log y}{\log a}
$$
by definition of $\log_a$. Therefore
$$
\log_{e^2}16=\frac{\log 16}{\log e^2}=\frac{2\log 4}{2}=\log 4
$$
and finally
$$
e^{\log_{e^2}16}=e^{\log 4}=4
$$
