How do you solve $Log(x+2)+log(x-1)=4$ ?

Question

7600 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-22T08:03:10

$x~=99.51$

$log(x+2)+log(x-1)=4$

$hArrlog((x+2)(x-1))=4$

i.e. $(x+2)(x-1)=10^4=10000$

or $x^2+x-10002=0$

and using quadratic formula $x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=(-1+-sqrt(1+40008))/2$

Now, we cannot use $-$ sign as that makes $log(x+2)$ or $log(x-1)$ not possible.

Hence $x=(-1+sqrt(1+40008))/2~=(-1+200.02)/2=199.02/2=99.51$

How do you solve Log(x+2)+log(x-1)=4Log(x+2)+log(x-1)=4?