Question

How do you solve `4^( x + 2) = 20`?

Answer 1

I found: `x=(ln(20))/(ln(4))-2=0.16096`

Explanation:

I would first take the natural log of both sides:
`ln(4)^(x+2)=ln(20)`
then use a property of logs to write:
`(x+2)ln(4)=ln(20)`
rearrange:
`x+2=(ln(20))/(ln(4))`
and:
`x=(ln(20))/(ln(4))-2=0.16096`