<PRE><B>Question 267189</B>
{{{log3 (x + 1) + log3 (x - 5) = 3}}}
Adding logs is the same as multiply
{{{log3((x + 1)*(x - 5)) = 3}}}
FOIL
{{{log3(x^2 - 5x + x - 5) = 3}}}
:
{{{log3(x^2 - 4x - 5) = 3}}} 
The exponent equiv of logs
{{{x^2 - 4x - 5 = 3^3}}}
:
x^2 - 4x - 5 = 27
:
x^2 - 4x - 5 - 27 = 0
:
x^2 - 4x - 32 = 0 
Factor
(x-8)(x+4) = 0
positive solution (cannot have a log of a neg number)
x = 8
:
See if this works
{{{log3 (9) + log3 (3) = 3}}}
{{{log3 (27) = 3}}}
Exponent equiv:
27 = 3^3</PRE>