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

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Answer 1 · 2018-05-28T11:11:45

$y=1/(1-e^x)$

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

Answer 2 · 2018-05-28T11:16:43

$y=(1/(1-e^x))$

Using Property of logarithm

$ln(a/b)=ln(a)-ln(b)$
$ln(y-1)-ln(y)=x$

$ln((y-1)/y)=x$

Now take antilog
$((y-1)/y)=e^x$

Simplify above equation
$y=(1/(1-e^x))$