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I want to prove some of the properties of logarithms.

a) $Log_{a} (x^{n})$ = $nlog_{a} x$

b) $log_{b^{n}} x$ = $\frac {1}{n} log_{b} x$

I have already proven then while making the post, but I thought that there might be some other ways to do it. Anyway here is how I proved them:

Demonstration

a)$$Log_{a} (x^{n}) = nlog_{a} x$$

I will use one of the properties of logarithms, which can also be proved, to demonstrate both properties. ($log_{a} x= \frac{log_{c} x}{log_{c}a}$)

So, here first I´ll use the concept of logarithms.

let $$log_{a}x=b $$ and $$Log_{a} (x^{n})=c $$

So $a^{b}=x$ and $a^{c}=x^{n}$

So:

$$a^{c}=x^{n}$$

$$ a^{c}=(a^{b})^{n}$$

So: $c =bn$ and we´d have:

$$Log_{a} (x^{n})=c $$

$$Log_{a} (x^{n})=bn$$

$$Log_{a} (x^{n})= nlog_{a}x$$

b) $$log_{b^{n}} x = \frac {1}{n} log_{b} x$$

$$log_{b^{n}} x = \frac {log_{c}x}{log_{c} (b^ {n})}$$

Using the property we have just show, we would have:

$$ = \frac {log_{c}x}{nlog_{c} b}$$

$$ = \frac {1}{n} (\frac{log_{c}x}{log_{c} b})$$

And applying the initial property we´d have:

$$log_{b^{n}} x = \frac {1}{n} log_{b} x$$

Is there any other way to prove them. Anyway, thanks in advance.

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If your definition of logarithm is something like $p=\log_q(r) \iff q^p=r$, and you use the usual properties of powers $(b^n)^{z}= b^{nz}$ and $\sqrt[n]{a^{y}}=a^{y/n}$ and further assume $x,a,b$ positive and $a,b$ not $1$ and $n$ not $0$, then

a) ${y} = \log_a(x^n) \iff a^{y}= x^n \iff a^{y/n} = x \iff \frac y n = \log_{a} (x) \iff y = n\log_{a} (x) $

b) $z = \log_{b^{n}} (x) \iff (b^n)^{z}= x \iff b^{nz}= x \iff nz= \log_{b}(x) \iff z= \frac1n\log_{b}(x)$

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  • $\begingroup$ Incredible! Thanks. $\endgroup$ – Vmimi Sep 8 '17 at 23:26
  • $\begingroup$ Is that definiton correct? $\endgroup$ – Vmimi Feb 2 '18 at 5:03
  • $\begingroup$ @joseestebanbeleñobarrozo Wikipedia says The logarithm of a positive real number $x$ with respect to base $b$, a positive real number not equal to $1$, is the exponent by which $b$ must be raised to yield $x$; in other words, the logarithm of $x$ to base $b$ is the solution $y$ to the equation $b^y=x$. This is indeed the typical definition $\endgroup$ – Henry Feb 2 '18 at 9:11

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