SOLUTION: Please help me solve this equation: {{{log(7,(x+4)) + log(7,(x-2)) = 1}}}, {{{x>2}}}

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Please help me solve this equation: {{{log(7,(x+4)) + log(7,(x-2)) = 1}}}, {{{x>2}}}      Log On


   

Question 141324: Please help me solve this equation:
log%287%2C%28x%2B4%29%29+%2B+log%287%2C%28x-2%29%29+=+1, x%3E2
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me solve this equation:
log%287%2C%28x%2B4%29%29+%2B+log%287%2C%28x-2%29%29+=+1, x%3E2


Use this rule on the left side:
               log%28B%2CA%29+%2B+log%28B%2CC%29=log%28B%2C%28AC%29%29

log%287%2C%28%28x%2B4%29%28x-2%29%29%29+=+1, x%3E2

Now use this rule to rewrite the equation:
               log%28B%2CA%29=C is equivalent to B%5EC=A%29

7%5E1=%28x%2B4%29%28x-2%29, x%3E2

Simplify, FOIL out the right side:

7=x%5E2-2x%2B4x-8, x%3E2 

7=x%5E2%2B2x-8, x%3E2

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

0=x%5E2%2B2x-8-7, x%3E2

0=x%5E2%2B2x-15, x%3E2

Factor the right side:

0=%28x-3%29%28x%2B5%29, x%3E2
             
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%3E2

So the only solution is x=3

Edwin