Question 472890
solve the equation
log2(x+2) + log2(x-4) = 4 
Same as:
{{{log2((x+2)*(x-4))}}} = 4
FOIL
{{{log2(x^2-2x-8)}}} = 4
exponent equiv of logs
x^2 - 2x - 8 = 2^4
:
x^2 - 2x - 8 = 16
:
x^2 - 2x - 8 - 16 = 0
:
x^2 - 2x - 24 = 0
Factors to
(x-6)(x+4) = 0
Two solutions
x = 6 
x = -4
:
x = 6 is the only solution to the original equation (can't have the log of a neg)