Solve for y if ln(y-1)=X+lny?

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Solve for y if ln(y-1)=X+lny?

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Answer 1 · 2018-05-28T11:11:45

$y = \frac{1}{1 - e^{x}}$ $y=1/(1-e^x)$

Explanation:

We have
$ln (y - 1) - ln (y) = x$ $ln(y-1)-ln(y)=x$
so
$ln (\frac{y - 1}{y}) = x$ $ln((y-1)/y)=x$
$\frac{y - 1}{y} = e^{x}$ $(y-1)/y=e^x$
$\frac{11}{y} = e^{x}$ $11/y=e^x$
$1 - e^{x} = \frac{1}{y}$ $1-e^x=1/y$
so
$y = \frac{1}{1 - e^{x}}$ $y=1/(1-e^x)$

Answer 2 · 2018-05-28T11:16:43

$y = (\frac{1}{1 - e^{x}})$ $y=(1/(1-e^x))$

Explanation:

Using Property of logarithm

$ln (\frac{a}{b}) = ln (a) - ln (b)$ $ln(a/b)=ln(a)-ln(b)$
$ln (y - 1) - ln (y) = x$ $ln(y-1)-ln(y)=x$

$ln (\frac{y - 1}{y}) = x$ $ln((y-1)/y)=x$

Now take antilog
$(\frac{y - 1}{y}) = e^{x}$ $((y-1)/y)=e^x$

Simplify above equation
$y = (\frac{1}{1 - e^{x}})$ $y=(1/(1-e^x))$

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Solve for y if ln(y-1)=X+lny?

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