Question

Correct one: Solve for x, `log_2 (log_3x) = 2`?

Answer 1

I got the same answer as the other answerer, using a different method: `x=64`

Explanation:

Start with `log_2(log_3x)=2`

Raise both sides to the power of 2:

`(log_2(log_3x))^2=2^2=4`

Note that `log_2("anything")^2`= `"anything"`

That is, `log_2(3)^2=3`, `log_2(6)^2=6` and so on.

So then `log_3(x)=4`

Now raise both sides to the power 3:

`log_3(x)^3=4^3=64`

Same as above, `log_3(x)^3=x`

So `x=64`.