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I need to prove the following: $$\log(x^{\frac{n}{m}}) = \frac{n}{m}\log(x)$$

In order to prove it I can only use the definition: $$\log(x) = \int_1^x{\frac{1}{t}dt}$$

I tried using some change of variables, but I was not able to get to the result. Can anybody help me?

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  • $\begingroup$ I would say, first use your allowed definition to prove $\log(ab) = \log a + \log b$. $\endgroup$ – GEdgar Apr 8 at 21:45
  • $\begingroup$ @GEdgar I already proved that in a previous exercise, but I don’t see how can I use it $\endgroup$ – Facu50196 Apr 8 at 21:46
  • $\begingroup$ Next, use induction to prove $\log(x^n) = n\log(x)$ for $n=1,2,3,4,\dots$. $\endgroup$ – GEdgar Apr 8 at 21:48
  • $\begingroup$ Yes, I had already done that. But I don’t know how to mix those things since induction is only used for natural numbers, and the exercise establishes that both m and n are Integer numbers $\endgroup$ – Facu50196 Apr 8 at 21:51
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If you know that $$\log{(x^n)}=n\log{(x)}$$ For $n\in\mathbb{N}$ then we can say $$m\log{(x^{\frac{n}m})}=\log{\left((x^{\frac{n}m})^m\right)}=\log{(x^n)}=n\log{(x)}$$ $$\therefore \log{(x^{\frac{n}m})}=\frac1m (m\log{(x^{\frac{n}m})})=\frac1m(n\log{(x)})=\frac{n}m\log{(x)}$$

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  • $\begingroup$ The OP is only allowed to use the integral definition of the log function. $\endgroup$ – John Douma Apr 8 at 22:12
  • $\begingroup$ The OP states in a comment "I have already done that" when asked about proving that $\log{(x^n)}=n\log{(x)}$ $\endgroup$ – Peter Foreman Apr 8 at 22:14
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Given (defined): $$\log(x^{m/n}) = \int_1^{x^{m/n}}{\frac{1}{t}dt}$$ where $n \ne 0$

Let $\displaystyle s = t^\frac{n}{m}$ $$\implies ds = \frac{n}{m}t^{\frac{n}{m}-1} dt$$ $$\implies ds = \frac{n}{m}\frac{s}{t} dt $$ $$\frac{m}{n}\frac{t}{s}ds = dt$$

Under the substitution and the new limits, we therefore get:

$$\Rightarrow \log(x^{m/n}) = \frac{m}{n}\int_1^x\frac{1}{t}\frac{t}{s}ds = \frac{m}{n}\int_1^x\frac{1}{s}ds = \frac{m}{n}\log(x)$$


Note that this can be extended to any real number $r \in \mathbb{R}$ and $r \ne 0$

Let $\displaystyle s = t^\frac{1}{r}$ $$\implies ds = \frac{1}{r}t^{\frac{1}{r}-1} dt$$ $$\implies ds = \frac{1}{r}\frac{s}{t} dt $$ $$r\frac{t}{s}ds = dt$$

Under the substitution and the new limits, we therefore get:

$$\Rightarrow \log(x^r) = r\int_1^x\frac{1}{t}\frac{t}{s}ds = r\int_1^x\frac{1}{s}ds = r\log(x)$$

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    $\begingroup$ Probably not the point of the exercise, but I would like to point out that this does not work for $n = 0$, although the formula holds. The proof in this case is very trivial however. $\endgroup$ – Jarne Renders Apr 8 at 22:21
  • $\begingroup$ I'll make a note for that. $\endgroup$ – user1952500 Apr 8 at 22:22

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