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:


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



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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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