Question

Question #0be98

Answer 1

`x~~3.53 or x~~-4.53`

Explanation:

`log(x+3)+log(x-2)=1`

Using the addition rule, we get:
`log(x^2+x-6)=1`

`10^(log(x^2+x-6))=10^1=10`

`x^2+x-6=10`

`x^2+x-16=0`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-1-sqrt(1^2-4(-16)))/2=(-1-sqrt(65))/2~~3.53`

`x=(-1+sqrt(1^2-4(-16)))/2=(-1+sqrt(65))/2~~-4.53`

Answer 2

`x=-1/2+sqrt65/2~~3.53`

Explanation:

`log(x+3)+log(x-2)= 1`
`log[(x + 3)(x-2)]=1`
`x^2+x-6=10`
`x^2+x+(1/2)^2=10+6+1/4`
`(x+1/2)^2= 65/4`
`x+1/2=+-sqrt65/2`
`x=-1/2+-sqrt65/2`
`x~~ -4.53, 3.53` => reject the negative root, since it's not in the domain and makes the equation undefined thus it's an "Extraneous"
root.
therefore:
`x~~3.53` => the only valid solution