Question

How do you solve `log_2(x+2) + log_2(x+6)=5`?

Answer 1

`x = 2`

Explanation:

Use `log a + log b = log ab` and, if #a^x=b, log_a(a^x)=x#,

`log_2 (x+2)+log_2(x+6)=log_2((x+2)(x+6))=5`, with `x+2 >0 and x+5 >0`. It is sufficient that `x> -2`.

So, `(x+2)(x+6)=2^5=32`
`x^2+8x-20=0`.
The roots are `x = 2` and `x = -10`.
As `x> -2`, the root `-10` is inadmissible. So, `x = 2`.