Log laws proof using only rational exponents

Question

For all real $a>0$ and rational $b>0$,

Show that $\ln(a^b)=b\ln(a)$

Answer 1

$b\ln a=b\int_1^a\frac{1}{x}dx=\int_1^a\frac{b}{x}dx$

Let $u=x^b$

Then $du= bx^{b-1}dx$ and $\frac{du}{u}=\frac{b}{x}dx$

Therefore, $\int_1^a\frac{b}{x}dx = \int_1^{a^b}\frac{du}{u}=\ln {a^b}$

Answer 2

$$x = \log_b(a^n)$$ $$b^x=a^n$$ $$(b^x)^{1/n}=(a^n)^{1/n}$$ $$b^{x/n}=a$$ $$x/n = \log_b(a)$$ $$x = n \log_b(a)$$ $$\log_b(a^n) = n\log_b(a)$$

Answer 3

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.

Answer 4

$\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)=... \\ $

Answer 5

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.