1
$\begingroup$

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

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

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, BLAZE, Community, graydad, colormegone
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ How exactly do you define $\ln$? Are you aware $e^{\ln (x)}=x,\forall x>0$? $\endgroup$ – user236182 Nov 9 '15 at 23:40
  • $\begingroup$ @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 $x>0$ $\endgroup$ – Yeah.. Nov 9 '15 at 23:52
  • $\begingroup$ @Yeah.. Then STATE that in your question so that users can answer your question properly, not everyone even reads the comments when they answer. $\endgroup$ – BLAZE Nov 10 '15 at 10:20
4
$\begingroup$

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

Let $u=x^b$

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ See the comments for OP's definition of $\ln$. $\endgroup$ – user236182 Nov 9 '15 at 23:54
  • $\begingroup$ @user236182 ah dang... there goes all my beautiful work proving this just using exponent laws. Oh well :P $\endgroup$ – Brevan Ellefsen Nov 9 '15 at 23:55
  • $\begingroup$ 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 $e^{b\ln (a)}=\left(e^{\ln(a)}\right)^b$. $\endgroup$ – user236182 Nov 9 '15 at 23:57
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ the OP wants a solution in terms of integrals I believe $\endgroup$ – Brevan Ellefsen Nov 10 '15 at 0:13
  • 1
    $\begingroup$ @BrevanEllefsen I don't think so. $\endgroup$ – egreg Nov 10 '15 at 0:13
  • $\begingroup$ @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. $\endgroup$ – BLAZE Nov 10 '15 at 3:49
0
$\begingroup$

$\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)=... \\ $

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

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.

$\endgroup$
  • $\begingroup$ OP wants a proof with integrals; see the comments. $\endgroup$ – user236182 Nov 10 '15 at 3:24
  • $\begingroup$ @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. $\endgroup$ – BLAZE Nov 10 '15 at 3:28

Not the answer you're looking for? Browse other questions tagged or ask your own question.