Question 626403: Find the exact solution, using common logarithms, and a two-decimal-place approximation, when appropriate.
log(x^(2)+6)-log(x+1)=2+log(x-1)
x=
x~
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the exact solution, using common logarithms, and a two-decimal-place approximation, when appropriate.
log(x^(2)+6)-log(x+1)=2+log(x-1)
log(x^(2)+6)-log(x+1)-log(x-1)=2
log(x^(2)+6)-[log(x+1)+log(x-1)]=2
place under single log
log[(x^2+6)/((x+1)(x-1))]=2
convert to exponential form
10^2=(x^2+6)/((x+1)(x-1))=100
(x^2+6)/((x+1)(x-1))=100
(x^2+6)/(x^2-1)=100
x^2+6=100x^2-100
99x^2-106=0
x^2=106/99
x=±√(106/99)
x≈-1.04(reject, x>1)
or
x≈1.04
|
|
|
