did I solve this righ?

log 2 + 3 log (X) = log 8
2 2 2

log 2Xraised to the 3rd = log 8
2 2
2X raised to the 3rd = 8
x raised to the 3rd = 8
x= squared root of 4


can anyone tell me if I did this right?

If I read what you wrote correctly I think your problem is to solve for X given this equation:
\log_2 2 + 3 \log_2 X = \log_2 8

The following identities will help:
\begin{eqnarray}
(1)&\log_b m + \log_b n &=& \log_b \left(m*n\right)\\
(2)&n \log_b m &=& \log_b \left(m^n\right)\\
(3)&b^{\log_b m}&=&m\\
\end{eqnarray}


Now we can begin to solve:
\log_2 2 + 3 \log_2 X = \log_2 8\\
\text{using identity (2):}\\
\log_2 2 + \log_2 X^3 = \log_2 8\\
\text{using identity (1):}\\
\log_2 {2*X^3}=\log_2 8\\
\text{make each side a the power of 2}\\
2^{\log_2 {2*X^3}}=2^{\log_2 8}\\
\text{using identity (3):}\\
2*X^3=8\\
\text{then solve for X:}\\
X^3=4\\
X=4^{\left(\frac 1 3\right)}\approx 1.5874\\


Comparing this to what you did it looks like you worked the problem correctly. The only concern I have is your final answer:

x= squared root of 4
The answer is:
x=cube root of 4
not square root of 4.
I suspect that's what you meant to write since you got everything else correct up to that point so good work!

Let us know if you have any additional questions.