Question

How do you solve `log_2(x+7)+ log_2 (x+8)=1`?

Answer 1

I found `x=-6`

Explanation:

You can use the property of the logs that tells us that:
`logx+logy=log(xy)`
and write:
`log_2[(x+7)(x+8)]=1`
now you apply the definition of log to change it into an exponential:
`[(x+7)(x+8)]=2^1`
`[(x+7)(x+8)]=2`
so that:
`x^2+8x+7x+56=2`
`x^2+15x+54=0`
We can use the Quadratic Formula:
`x_(1,2)=(-15+-sqrt(225-216))/2=(-15+-3)/2`
Two solutions:
`x_1=-9` NO it makes both argument negative.
`x_2=-6`