How do you solve $Ln(x+1)-ln(x-2)=lnx^2$ ?

Question

How do you solve $Ln(x+1)-ln(x-2)=lnx^2$ ?

1 Answer

Impact of this question

1811 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-10T04:29:45

approximately:
$x=2.5468$

Explanation:

$ln^[(x+1)/(x-2)]$ = $ln^(x^2)$
we can cancel out the (Ln) parts and the exponents would be left out;
$(x+1)/(x-2)$ = $x^2$
$x+1$ = $x^2.(x-2)$
$x+1$ = $x^3-2x^2$
$x^3$ - $2x^2$ - $x$ - $1$ = $0$
$x=2.5468$

Science

Math

Humanities

... and beyond

How do you solve $Ln(x+1)-ln(x-2)=lnx^2$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve Ln(x+1)-ln(x-2)=lnx^2Ln(x+1)-ln(x-2)=lnx^2?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $Ln(x+1)-ln(x-2)=lnx^2$ ?