SOLUTION: Need to know how to solve ln(x/3+1)+ln(1/x)=2

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Question 958171: Need to know how to solve ln(x/3+1)+ln(1/x)=2
Found 2 solutions by Alan3354, rothauserc:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Need to know how to solve ln(x/3+1)+ln(1/x)=2
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Adding logs --> multiply
ln%28%28x%2F3+%2B+1%29%2Fx%29+=+2
ln%28%28x%2B3%29%2F3x%29+=+2
%28%28x%2B3%29%2F3x%29+=+e%5E2
x%2B3+=+3e%5E2%2Ax
3+=+3e%5E2%2Ax+-x+=+x%2A%283e%5E2+-+1%29
x+=+3%2F%283e%5E2+-+1%29

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
ln(x/3+1)+ln(1/x)=2
assuming x is positive, use logarithm product rule and we have
ln ( (x/3 + 1) * (1/x) ) = 2
ln (1/3 + 1/x ) = 2
use definition of logarithms
1/3 + 1/x = e^2
1/x = e^2 - 1/3
1/x = (3e^2 - 1) / 3
cross multiply fractions
(3e^2 - 1)*x = 3
x = 3 / (3e^2 - 1)