Question

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

Answer 1

`x=-4 " or " +3`

Explanation:

Let "log base" be b

Known that `log_b(b)=1`

Assumption: The use of 'log' in the question is referring to logs to base 10 giving:

`log_10(x+2) +log_10(x-1) =1.....................(1)`

But `color(white)(xxxxxx)log_10(10)=1 .............................(2)`

Substitute (2) into (1) giving:

`log_10(x+2)+log_10(x-1) = log_10(10) ..........(3)`

But `log_10(x+2)+log_10(x-1) = log_10[ (x+2)(x-1)]`

Rewrite (3) as:

`log_10[(x+2)(x-1)] = log_10(10)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Personal note: I am of 'very' old school where `"antilog" -> log^(-1)`

Taking antilogs `-> (x+2)(x-1)=10`

`=> x^2+x-12=0`

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

`color(blue)( x=-4 " or " +3)`