I'm assuming the LHS is [log(x^2)]^2 and the RHS is log(x^4)?

Then

[2 log(x)]^2 = 4 log(x)

Set y = log(x)

[2y]^2 = 4y

4y^2 = 4y

y^2 = y

y^2 - y = 0

y(y - 1) = 0

So y = 0 or y = 1.

Thus log(x) = 0 or log(x) = 1

log(x) = 0

10^[log(x)] = 10^0

x = 1

and

log(x) = 1

10^[log(x)] = 10^1

x = 10.

Thus x = 1 or x = 10.

-Dan