Question

How do you solve `2^(x+4)=1/32`?

Answer 1

Consider that you can write: `1/32=1/2^5=2^-5`
Your equation becomes:
`2^(x+4)=2^-5`
Take the `log_2` of both sides and remember that `log_aa^x=x`;
So you get:
`log_2(2^(x+4))=log_2(2^-5)`
`x+4=-5`
`x=-9`