SOLUTION: i have the problem: ln(x+1)^2=2 i worked it out as: ln(x+1)^2=2 (x+1)^2=e^2 2(x+1)=7.38906 2x+2=7.38906 2x\2 = 5.38906 x=2.69453 i know i did something wrong becaus

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: i have the problem: ln(x+1)^2=2 i worked it out as: ln(x+1)^2=2 (x+1)^2=e^2 2(x+1)=7.38906 2x+2=7.38906 2x\2 = 5.38906 x=2.69453 i know i did something wrong becaus      Log On


   

Question 209720: i have the problem:
ln(x+1)^2=2
i worked it out as:
ln(x+1)^2=2
(x+1)^2=e^2
2(x+1)=7.38906
2x+2=7.38906
2x\2 = 5.38906
x=2.69453

i know i did something wrong becausemy answer is not working out.
help me plz!
what did i do wrong?
thanks in advanced
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
ln(x+1)^2=2
i worked it out as:
ln(x+1)^2=2
(x+1)^2=e^2
2(x+1)=7.38906
2x+2=7.38906
2x\2 = 5.38906
x=2.69453
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When you have ln(x+1)^2 = 2 you get the following:
2*ln(x+1) = 2
ln(x+1) = 1
x+1 = e
x = 2-1
x = 1.71828....
==========================
Cheers,
Stan H.
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