Question 398624
{{{log(b,x^2 -15) = log(b,6x + 1)}}}
Suppose I say 
{{{c = x^2 - 15}}} and
{{{d = 6x + 1}}}
Then I can rewrite equation as:
{{{log(b,c) = log(b,d)}}}
The main thing to keep in mind is "logs are exponents"
This equation says:
The log to the base {{{b}}} which gives me {{{c}}} is equal to
the log to the base {{{b}}} which gives me {{{d}}}
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This is 2 logs equal to each other, so I can say that log =  {{{e}}}
{{{b^e = c}}} is true
{{{b^e = d}}} is true
So, {{{c = d}}} must be true
and, since I said
{{{c = x^2 - 15}}} and
{{{d = 6x + 1}}}
It follows
{{{x^2 - 15 = 6x + 1}}}
{{{x^2 - 6x - 16 = 0}}}
By looking at it, I see
{{{(x - 8)*(x + 2)}}}
The solutions are:
{{{x = 8}}} and
{{{x = -2}}}