Logarithm rules and how they work as algebra I am having trouble understanding how logarithms work. If 
$\log_a x = n$ that is the same as $a^n = x$ 
So if that is true than the log rule
$\log_b(x^n) = n \log_b x$ is the same as $a^z = x^n$ which is the same as $n(a^z) = x$
But using numbers for this it is not true. What went wrong?
 A: Indeed, $\log_b(r)=s$ means that $b^s=r$.
So, let's look at $\log_b(x^n)$ and $n\log_b(x)$. Let $t=\log_b(x^n)$; this means that $b^t = x^n$. And let $u = \log_b(x)$. That means that $b^u = x$.
Then $n\log_b(x) = nu$. Now, what is $b^{nu}$? 
$$b^{nu} = b^{un} = \left(b^u\right)^n = x^n = b^t$$
(because $b^u = x)$. So, indeed, $nu=n\log_b(x)$ is the same thing as $t=\log_b(x^n)$. 
Every rule of logarithms corresponds to a rule of exponentiation. Here, the rule that says that $\log_b(x^n) = n\log_b(x)$ corresponds to the rule of exponentiation that says that
$$(b^r)^s = b^{rs}.$$
For another example, the rule that says that $\log_b(xy) = \log_b(x) + \log_b(y)$ corresponds to the exponentiation rule that says that $b^rb^s = b^{r+s}$. Indeed, say $r=\log_b(x)$ and $s=\log_b(y)$. Then $b^r=x$, $b^s=y$, so $xy = (b^r)(b^s) = b^{r+s}$, so $\log_b(xy) = r+s = \log_b(x)+\log_b(y)$.
A: $\log_b(x^n) = n \log_b x$   is incorrect,it should be $\log_b(x^n)=n\log_b(x)$,so  in case of $a^z=x^n$   $\ln(a^z)=\ln(x^n)$  ,so  $z\ln a =n\ln x$ ,from here $\ln x=(z/n)\ln a$  and finally  $x=\exp((z/n)\ln a)$  or   $x=a^(z/n)$( i wanted to write a in power  of $z/n$)
