SOLUTION: How do you solve the following equations? log2(3x+2)-log2(x+2)=1 and ln(x-1)=4-ln(x+2) i've tried repeatedly and my professor won't help!

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Question 64162: How do you solve the following equations?
log2(3x+2)-log2(x+2)=1

and
ln(x-1)=4-ln(x+2)

i've tried repeatedly and my professor won't help!
Found 2 solutions by josmiceli, stanbon:
Answer by josmiceli(6783) About Me  (Show Source):
Answer by stanbon(48535) About Me  (Show Source):
You can put this solution on YOUR website!
How do you solve the following equations?
log2(3x+2)-log2(x+2)=1
log2[(3x+2)/(x+2)]=1
[(3x+2)/(x+2)]=2^1
3x+2=2(x+2)
3x+2=2x+4
x=2
----------------

ln(x-1)=4-ln(x+2)
ln(x-1)+ln(x+2)=4
ln[(x-1)(x+2)]=4
(x-1)(x+2)=e^4
x^2+x-2=54.598
x^2+x-56.598=0
x=[-1+-sqrt(1-4*-56.598)]/2
x=[-1+-sqrt(227.392)]/2
x=7.03976...
The negative answer will not work in the original equation.
Cheers,
Stan H.