Hello again AeroScizor Originally Posted by

**AeroScizor** ...

3) Solve x in terms of a given that log base a 5 + 2= log base a (x+a) + log base a (x-3a).

Help would be appreciated. Thanks in advance.

Write $\displaystyle 2$ as $\displaystyle 2\log_a(a)=\log_a(a^2)$, and then:

$\displaystyle \log_a5+\log_a(a^2) = \log_a(x+a)+\log_a(x-3a)$

$\displaystyle \Rightarrow \log_a(5a^2)=\log_a(x+a)(x-3a)$

$\displaystyle \Rightarrow 5a^2 = ...$

Simplify, factorise and solve the quadratic in $\displaystyle x$. Not both roots are valid though, because $\displaystyle a,\; (x+a)$ and $\displaystyle (x-3a)$ must all be positive.

Answer: $\displaystyle x = 4a$

Can you complete it now?

Grandad