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

Question

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

1 Answer

Impact of this question

1693 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-16T11:20:51

From laws of logs we get

$ln [\frac{x + 1}{x - 2}] = {ln x}^{2}$ $ln[(x+1)/(x-2)]=lnx^2$

$therefore (x+1)/(x-2)=x^2$

$therefore x^3-2x^2-x-1=0$

This is now a 3rd degree polynomial so you will have to try using the factor and remainder theorem to find the roots, otherwise you will have to use a numerical technique such as Newton's method of root-finding.

I leave the details as an exercise. Please ask if you are still stuck.

Science

Math

Humanities

... and beyond

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

1 Answer

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)=lnx2ln(x+1)-ln(x-2)=lnx^2?

1 Answer

Related questions

Impact of this question

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