Question

How do you solve for `log_9 27 = x`?

Answer 1

It is

`x=log_9 27=log_9 3^3=3*log3/log9=3*(log3/log3^2)=3/2`

Finally `x=3/2`

Answer 2

I found: `x=3/2`

Explanation:

We can use the definition of log:
`log_bx=a->x=b^a`
and write:
`27=9^x` that we can write as:

`3^3=3^(2x)`
for the two terms to be equal also the exponents must be equal, so:
`3=2x`
and:
`x=3/2`