Prove $\log_{a} c + \log_{b} c = \log_{a+b} c$ if and only if $1 + \log_{b} a = \log_{a+b} a$

Question

If a, b and c are positive numbers, than equality $$\log_{a} c + \log_{b} c = \log_{a+b} c$$ is true if and only if $$1 + \log_{b} a = \log_{a+b} a$$ Prove it!

I have looked at the solution but it is not clear for me.

We will prove that if $$\log_{a} c + \log_{b} c = \log_{a+b} c$$ than it is $$1 + \log_{b} a = \log_{a+b} a$$

$$\log_{a} c + \log_{b} c = \frac{\log_{a+b} c}{\log_{a+b} a} +\frac{\log_{a+b} c}{\log_{a+b} b} =\log_{a+b} c $$

So this is only thing that is not clear for me how is this equal. $$\frac{\log_{a+b} c}{\log_{a+b} a} +\frac{\log_{a+b} c}{\log_{a+b} b} =\log_{a+b} c $$

Answer 1

Note that

$$log_b (a)= \frac{ln (a)}{ln(b)}$$ where $ln(a)$ is the natural log of $a$.

Any arbitrary base can be assumed instead of $e$.

So, if

$$log_a (c)+ log_b (c)= log_{a+b} (c) $$

$$\implies\frac{ln(c)}{ln(a)}+\frac{ln(c)}{ln(b)}=\frac{ln(c)}{ln(a+b)}$$ $$\implies\frac{1}{ln(a)}+\frac{1}{ln(b)}=\frac{1}{ln(a+b)}\ ,\ \ assuming \ \ c\neq1$$ $$\implies1+\frac{ln(a)}{ln(b)}=\frac{ln(a)}{ln(a+b)}$$ $$\implies1+log_b(a)=\frac{ln(a)}{ln(a+b)}$$ $$\implies1+log_b(a)=log_{a+b}(a)$$

The steps are perfectly reversible.

Answer 2

WE have:

$$1+ log_b (a)= log_{a+b} (a) $$

Let $a=c$ then:

$$ log _a(c)= log _c (c)=1$$

therefore:

$$log _a (c) + log _b (c)= log _{a+b} (c)$$