How to show that: $\log_a (x^{a}-x)-\log_a \Big(\dfrac{x^{a}-x}{a}\Big)=1$, where $a$ and $x$ are positive integers.

Question

I was studying Fermat's Little Theorem and Logarithm to see if there is any interesting result or correlation exist between the two. So I came up with this equation. I know few basic logarithmic properties and identities but none seem to help to give any insight or answer. May be the result is trivial but I'm confused, so I need some help.

Regards

Answer 1

Use:

1) $\log_a X - \log_a Y=\log_a \frac{X}Y$

2) $\log_a a=1$

$$\log_a (x^{a}-x)-\log_a \Big(\dfrac{x^{a}-x}{a}\Big)=$$ $$=\log_a\frac{(x^{a}-x)}{\Big(\dfrac{x^{a}-x}{a}\Big)}=\log_a a=1$$

Answer 2

Use:

• $$\log_a(b)=\frac{\ln(b)}{\ln(a)}$$
• When $a$ and $b$ are positive: $$\ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)$$

So, we get when $a$ and $x$ are positive ($x\ne1$ and $a\ne1$):

$$\log_a\left(x^a-x\right)-\log_a\left(\frac{x^a-x}{a}\right)=\frac{\ln\left(x^a-x\right)}{\ln(a)}-\frac{\ln\left(x^a-x\right)-\ln\left(a\right)}{\ln(a)}=$$ $$\frac{\ln\left(x^a-x\right)-\ln\left(x^a-x\right)+\ln\left(a\right)}{\ln(a)}=\frac{0+\ln\left(a\right)}{\ln(a)}=\frac{\ln\left(a\right)}{\ln(a)}=1$$