SOLUTION: log6 (x+1)+log6(x)=1

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: log6 (x+1)+log6(x)=1      Log On


   

Question 136175: log6 (x+1)+log6(x)=1
Found 2 solutions by Fombitz, Lightning_Fast:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
log%286%2C%28x%2B1%29%29%2Blog%286%2C%28x%29%29=1
log%286%2C%28x%28x%2B1%29%29%29=1
6%5E%281%29=x%28x%2B1%29
x%5E2%2Bx=6
x%5E2%2Bx-6=0
%28x%2B3%29%28x-2%29=0
x=2 and x=-3.
Check your answers.
log%286%2C%28x%28x%2B1%29%29%29=1
log%286%2C%28-3%28-2%29%29%29=1
log%286%2C6%29=1
True statement, but only for the converted equation.
In your original equation,
log%286%2C%28x%2B1%29%29%2Blog%286%2C%28x%29%29=1
log%286%2C%28-2%29%29%2Blog%286%2C%28-3%29%29=1
The solution x=-3 is not allowed, since the logarithm function must have positive arguments.
log%286%2C%282%2B1%29%29%2Blog%286%2C%282%29%29=1
log%286%2C%283%29%29%2Blog%286%2C%282%29%29=1
log%2810%2C3%29%2Flog%2810%2C6%29%2Blog%2810%2C2%29%2Flog%2810%2C6%29=1
0.613147%2B0.386853=1
1=1
True statement.
Good answer.
x=2.

Answer by Lightning_Fast(78) About Me  (Show Source):
You can put this solution on YOUR website!
log6%28x%28x%2B1%29%29=1 --> log6%28x%5E2%2Bx%29=1 --> 6%5E1=x%5E2%2Bx --> 0=x%5E2%2Bx-6 --> %28x-2%29%28x%2B3%29=0
-3 is extraneous so the answer is
x=2