Question 141324
Please help me solve this equation:

{{{log(7,(x+4)) + log(7,(x-2)) = 1}}}, {{{x>2}}}
<pre><font size = 4 color="indigo"><b>

Use this rule on the left side:
               {{{log(B,A) + log(B,C)=log(B,(AC))}}}

{{{log(7,((x+4)(x-2))) = 1}}}, {{{x>2}}}

Now use this rule to rewrite the equation:
               {{{log(B,A)=C}}} is equivalent to {{{B^C=A)}}}

{{{7^1=(x+4)(x-2)}}}, {{{x>2}}}

Simplify, FOIL out the right side:

{{{7=x^2-2x+4x-8}}}, {{{x>2}}} 

{{{7=x^2+2x-8}}}, {{{x>2}}}

Get {{{0}}} on the left by subtracting
7 from both sides:

{{{0=x^2+2x-8-7}}}, {{{x>2}}}

{{{0=x^2+2x-15}}}, {{{x>2}}}

Factor the right side:

{{{0=(x-3)(x+5)}}}, {{{x>2}}}
             
Use the zero-factor principle:

x-3=0 gives x=3
x+5=0 gives x=-5

However we must discard that second answer
because we are given the restriction that {{{x>2}}}

So the only solution is {{{x=3}}}

Edwin</pre>