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Hoofbeat
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Hi, I have a few questions from P2 that I'm stuck with, so if anyone could help that would be great.

1. Find value of y to 2d.p:
ln(y+1) - lny = 0.85
Whenever I do it I end up with y's that cancel out! grrrr

2.Is e^(x-2) the same as e^x divided by e^2 (using indice rules)? Because I have a question where I have to find the inverse function of f(x)=e^(x-2) - 1 and I wasn't entirely sure what happens when I ln the equation, to get x.

Thanks ever so much
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Nylex
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(Original post by Hoofbeat)
Hi, I have a few questions from P2 that I'm stuck with, so if anyone could help that would be great.

1. Find value of y to 2d.p:
ln(y+1) - lny = 0.85
Whenever I do it I end up with y's that cancel out! grrrr

2.Is e^(x-2) the same as e^x divided by e^2 (using indice rules)? Because I have a question where I have to find the inverse function of f(x)=e^(x-2) - 1 and I wasn't entirely sure what happens when I ln the equation, to get x.

Thanks ever so much
1. ln (y + 1) - ln y = 0.85

=> ln |(y + 1)/y| = 0.85

(y + 1)/y = e^0.85

(y + 1)/y = 2.3396

2.3396y = y + 1

2.3396y - y = 1

y(2.3396 - 1) = 1

y = 1/1.3396
y = 0.75

2. Yes, it is.
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Hoofbeat
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Report Thread starter 16 years ago
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(Original post by Nylex)
1. ln (y + 1) - ln y = 0.85

=> ln |(y + 1)/y| = 0.85

(y + 1)/y = e^0.85

(y + 1)/y = 2.3396

2.3396y = y + 1

2.3396y - y = 1

y(2.3396 - 1) = 1

y = 1/1.3396
y = 0.75

2. Yes, it is.
You're a star!!!!!!!!!! (((HUGZ))) Thanks so much!
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Nylex
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No prob .
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