SOLUTION: ln e^(log(x+1)+log(x-2)) = (1/2)e^(ln(2)), what is x?

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: ln e^(log(x+1)+log(x-2)) = (1/2)e^(ln(2)), what is x?      Log On


   

Question 792749: ln e^(log(x+1)+log(x-2)) = (1/2)e^(ln(2)), what is x?
Found 2 solutions by psbhowmick, lwsshak3:
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
Can you type the equation properly... it is not very clear which are log to the base 10 and which are log to the base e.

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
ln e^(log(x+1)+log(x-2)) = (1/2)e^(ln(2)), what is x?
***
log of base raised to a power=power
ln e^(log(x+1)+log(x-2))=log(x+1)+log(x-2)=log(x+1)(x-2)
..
Base raised to the log of number=number
(1/2)e^(ln(2))=(1/2)*2=1
..
log(x+1)(x-2)=1
10^1=(x+1)(x-2)
10=x^2-x-2
x^2-x-12=0
(x-4)(x+3)=0
x=4
or
x=-3