Question 276770
I'm not sure how well you understand logs. You have to
know the general rules, The best rule to remember is
"logs are exponents". The rule that applies here is:
{{{log(a,(b/c)) = log(a,b) - log(a,c)}}}
Your problem:
{{{log(4,(x^2 + 3x)) - log(4,(x+5))}}}
Using the rule in reverse,
{{{log(4,((x^2 + 3x)/(x + 5)))}}}
In order to solve for {{{x}}}, you need this expression 
to equal some number, for instance, if it equals {{{3}}},
{{{log(4,((x^2 + 3x)/(x + 5))) = 3}}}
This is the same as
{{{4^3 = (x^2 + 3x)/(x+5)}}}
{{{64 = (x^2 + 3x)/(x+5)}}}
{{{64x + 320 = x^2 + 3x}}}
{{{x^2 - 61x - 320}}}
This can be solved using quadratic formula