# Log laws proof using only rational exponents [closed]

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

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

## closed as off-topic by user223391, BLAZE, user99914, graydad, colormegoneNov 10 '15 at 3:42

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• How exactly do you define $\ln$? Are you aware $e^{\ln (x)}=x,\forall x>0$? – user236182 Nov 9 '15 at 23:40
• @user236182 Personally, I know what $\ln$ is but for the purpose of this proof I want to define it as the area between $1$ and $x_0$ under the curve $\frac1x$ where $x>0$ – Yeah.. Nov 9 '15 at 23:52

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

Let $u=x^b$

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

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

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

• See the comments for OP's definition of $\ln$. – user236182 Nov 9 '15 at 23:54
• @user236182 ah dang... there goes all my beautiful work proving this just using exponent laws. Oh well :P – Brevan Ellefsen Nov 9 '15 at 23:55
• I still think it's much easier to just notice $\ln\left(a^b\right)=b\ln(a)\iff e^{\ln\left(a^b\right)}=e^{b\ln(a)}$ and $e^{b\ln (a)}=\left(e^{\ln(a)}\right)^b$. – user236182 Nov 9 '15 at 23:57

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.

• the OP wants a solution in terms of integrals I believe – Brevan Ellefsen Nov 10 '15 at 0:13
• @BrevanEllefsen I don't think so. – egreg Nov 10 '15 at 0:13
• @Brevan You're right, but the OP should have stated that in the question, on viewing the question I have also answered not using integrals. – BLAZE Nov 10 '15 at 3:49

$\ln(x)=\int_1^x \frac{1}{t} dt \\ \text{ so } \ln(x^r)=\int_1^{x^r} \frac{1}{t} dt \\ \text{ and then differentiating both sides gives} \\ [\ln(x^r)]'=(x^r)' \cdot \frac{1}{x^r}=\frac{r}{x} \\ \text{ now integrating both sides ... see if you can finish from here } \ln(x^r)=... \\$

• I see John's answer now. He went in the opposite direction of me. – randomgirl Nov 10 '15 at 0:27

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.

• OP wants a proof with integrals; see the comments. – user236182 Nov 10 '15 at 3:24
• @user236182 Thanks for pointing that out, I spotted that after I posted. But since the others have already posted good answers via integration I shall leave this as it is as I was simply answering the original question that was posted by the OP and it says nothing about using integrals in the question as you can see. – BLAZE Nov 10 '15 at 3:28