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I need to prove this logarithm.

$$\log_p\Big(\frac{1}{x}\Big) = -\log_p(x)$$

The first step would be $\ln(1/x)/\ln_p$

I need help as to what the next step would be.

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  • $\begingroup$ what is your first step¿ $\endgroup$
    – Surb
    Nov 19, 2015 at 21:19
  • $\begingroup$ ln(1/x) / ln_p is the first step $\endgroup$
    – Cheng
    Nov 19, 2015 at 21:20
  • $\begingroup$ ok, well then I don't understand it. $\endgroup$
    – Surb
    Nov 19, 2015 at 21:21
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    $\begingroup$ Can't really answer this question if you don't tell what we are allowed to use. $\endgroup$
    – user65203
    Nov 19, 2015 at 21:24
  • $\begingroup$ I just have to show that the identity holds $\endgroup$
    – Cheng
    Nov 19, 2015 at 21:25

5 Answers 5

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In any base,

$$\log(xy)=\log(x)+\log(y)$$ hence

$$\log(x)+\log(\frac1x)=\log(x\frac1x)=\log(1)=0.$$

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$\log_p\frac{1}{x} = \log_p1-\log_p x=-\log_p x$

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Assuming $x\in\mathbb{R}^+$:

$$\log_p\left(\frac{1}{x}\right)=\frac{\ln\left(\frac{1}{x}\right)}{\ln(p)}=\frac{-\ln(x)}{\ln(p)}=-\frac{\ln(x)}{\ln(p)}=-\log_p(x)$$

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One can define the natural logarithm as $$\ln(x)=\lim_{n\to\infty}n\left(\sqrt[n]x-1\right).$$

Then

$$\ln\left(\frac1x\right)=\lim_{n\to\infty}n\left(\frac1{\sqrt[n]x}-1\right)=\lim_{n\to\infty}n\left(\frac{1-{\sqrt[n]x}}{\sqrt[n]x}\right)= \frac{\lim_{n\to\infty}n\left(1-{\sqrt[n]x}\right)}{\lim_{n\to\infty}\sqrt[n]x}=-\ln(x).$$

This immediately extends to other bases.

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Using $\log\left( a^b \right) = b \cdot \log(a)$ we get $$ \log_p\left( \dfrac{1}{x} \right) = \log_p\left( x^{-1} \right) = -\log_p\left( x \right). $$

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