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 because
Question 209715: 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! Answer by Alan3354(69443) (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 You changed from exponent 2 to multiply by 2
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2(x+1)=7.38906
2x+2=7.38906
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ln(x+1)^2=2
2ln(x+1) = 2
ln(x+1) = 1
x+1 = e
x = e-1
