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

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.
