Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you solve $ln (x + 1) - 1 = ln (x - 1)$ $ln(x+1) - 1 = ln(x-1)$ ?

1 Answer

Feb 24, 2016

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

Explanation:

Given
$XXX ln (x + 1) - 1 = ln (x - 1)$ $color(white)("XXX")ln(x+1)-1=ln(x-1)$

$\Rightarrow$ $rArr$
$XXX ln (x + 1) - ln (x - 1) = 1 = ln (e)$ $color(white)("XXX")ln(x+1)-ln(x-1) = 1 = ln(e)$

$XXX ln (\frac{x + 1}{x - 1}) = ln (e)$ $color(white)("XXX")ln((x+1)/(x-1))= ln(e)$

$XXX \frac{x + 1}{x - 1} = e$ $color(white)("XXX")(x+1)/(x-1)=e$

$XXX x + 1 = e x - e$ $color(white)("XXX")x+1=ex-e$

$XXX x - e x = - e - 1$ $color(white)("XXX")x-ex=-e-1$

$XXX x (1 - e) = - e - 1$ $color(white)("XXX")x(1-e)=-e-1$

$XXX x = \frac{e + 1}{e - 1} \approx 2.164$ $color(white)("XXX")x=(e+1)/(e-1)~~2.164$

Related questions

See all questions in Logarithmic Models

Impact of this question

1365 views around the world

You can reuse this answer
Creative Commons License