Logs with exponential bases

Question

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.

Answer 1

$$\exp{\log_{e^2}(16)} = \exp\frac{\log{16}}{\log(e^2)} = \exp\frac{\log{16}}{2} = \exp{\log(16^{1/2})} = 16^{1/2} = 4.$$

I used the following rules: $\log_a(x) = \frac{\log(x)}{\log(a)}$, and $c\log(x) = \log(x^c)$.

Answer 2

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.

Answer 3

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!

Answer 4

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 $b$, $b^{\log_b(x)}=x$.

Answer 5

You can use the following formula for logarithms $$\log_{a^q}b=\tfrac1q \log_ab$$

Answer 6

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 $$ e^{\log_{e^2}16}=e^{\log 4}=4 $$