SOLUTION: 1.Solve and check. Show your work. log(x-2)+log(x+1)=2 log 2

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: 1.Solve and check. Show your work. log(x-2)+log(x+1)=2 log 2      Log On


   

Question 1066900: 1.Solve and check. Show your work. log(x-2)+log(x+1)=2 log 2
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
First, raise e to both sides:
+e%5E%28log%28%28x-2%29%29%2Blog%28%28x%2B1%29%29%29+=+e%5E%282%2Alog%282%29%29+
The left hand side becomes (x-2)(x+1)
The right hand side becomes +e%5E%28log%28%282%29%29%2A2%29+=+2%5E2+=+4+
So +%28x-2%29%28x%2B1%29+=+4+
+x%5E2+%2Bx+-2x+-2+=+4+
++x%5E2+-x+-+6+=+0+
++%28x-3%29%28x%2B2%29+=+0+

x=3 and x=-2.
--
Discard x=-2 because log(-2 - 2) = log(-4) is undefined.
—
Check x=3: log(3-2)+log(3+1) = log(1) + log(4) = 0 + log(4) = log(2^2) = 2log(2) (ok)
—
Answer: +highlight%28x=3%29+