Log laws proof using only rational exponents For all real $a>0$ and rational $b>0$,
Show that $\ln(a^b)=b\ln(a)$
 A: $b\ln a=b\int_1^a\frac{1}{x}dx=\int_1^a\frac{b}{x}dx$
Let $u=x^b$
Then $du= bx^{b-1}dx$ and $\frac{du}{u}=\frac{b}{x}dx$
Therefore, $\int_1^a\frac{b}{x}dx = \int_1^{a^b}\frac{du}{u}=\ln {a^b}$
A: $$x = \log_b(a^n)$$
$$b^x=a^n$$
$$(b^x)^{1/n}=(a^n)^{1/n}$$
$$b^{x/n}=a$$
$$x/n = \log_b(a)$$
$$x = n \log_b(a)$$
$$\log_b(a^n) = n\log_b(a)$$
A: I guess you have already proved the main property
$$
\ln(xy)=\ln x+\ln y
$$
(for positive reals $x$ and $y$). By an easy induction you get that
$$
\ln(a^m)=m\ln a
$$
for a positive integer $m$. If $m<0$, you have
$$
\ln(a^m)=\ln\frac{1}{a^{-m}}=-\ln(a^{-m})=-(-m\ln a)=m\ln a
$$
Now suppose $b=m/n$, where, without loss of generality, $n>0$. Then
$$
\ln(a^b)=\ln((a^{1/n})^m)=m\ln(a^{1/n})
$$
by the above argument, so we're left to prove that
$$
\ln(a^{1/n})=\frac{1}{n}\ln a
$$
Since
$$
n\left(\frac{1}{n}\ln a\right)=\ln a
$$
and
$$
n\ln(a^{1/n})=\ln((a^{1/n})^n)=\ln a
$$
we get the desired result.
A: $\ln(x)=\int_1^x \frac{1}{t} dt \\ \text{ so } \ln(x^r)=\int_1^{x^r} \frac{1}{t} dt \\ \text{ and then differentiating both sides gives} \\ [\ln(x^r)]'=(x^r)' \cdot \frac{1}{x^r}=\frac{r}{x} \\ \text{ now integrating both sides ... see if you can finish from here } \ln(x^r)=... \\ $
A: Assuming knowledge of 
$(a^m)^n=a^{mn}\tag{1}$
Now let $a^m=b$ and $a^n = c$
Then from $(1)$
we have $b^n=(a^m)^n=a^{mn}$
so $\log_a(b^n)=mn$ 
therefore $n\log_a b=\log_a(b^n)$
Hence 
$\color{blue}{\fbox{$\log_a (b^n) = n\log_a b$}}$
I did this for general logarithms, as I think it's good practice to generalize proofs, but the above proof works exactly the same for natural logarithms (which you were asking about) also. 
