Question

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

Answer 1

see below

Explanation:

Using the addition property of logs you know that
this equals log(x+2)(x-1)=1
and using the common log you know that (x+2)(x-1) needs to equal 10 for the equation to be true, (x+2)(x-1)=10 and
this evaluates to (x+4)(x-3)=0, but -4 can't be a solution because that would make one of the original logs undefined.

Answer 2

`log(x+2)+log(x−1)=1`

`=> log(x+2)(x−1)=log10` `color(white)(x=3` `["as " loga+logb = logab]`

`=> (x+2)(x−1)=10` `color(white)(xwwww3` `["taking anti log on both sides"]`

`=> x^2+2x-x−2=10`

`=> x^2+x−12=0`

`=> x^2+4x-3x−12=0`

`=> x(x+4)-3(x+4)=0`

`=> (x-3)(x+4)=0`

Therefore, `x=3` or `x=-4`

since, `x` can't be negative, `color(teal)(x=3`