Question

How do you solve `log_36x= 1/2`?

Answer 1

`x= 6`

Explanation:

`log_(36) x = 1/2`

Switch to exponential form of `log_a x= y <=> a^y = x`

`log_(36)x= 1/2`
`=> (36)^(1/2) = x`
`=> sqrt(36)= x`

=>`x= 6`