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