solve for \( x \). Leave answer in radical form. \( \log _{4} x+\log _{4} 2 x=3 \)

Solve the equation \( \log_{4}x+\log_{4}2x=3 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\log_{4}{\left(x\right)}+\log_{4}{\left(2x\right)}=3\) - step1: Find the domain: \(\log_{4}{\left(x\right)}+\log_{4}{\left(2x\right)}=3,x>0\) - step2: Add the terms: \(\log_{4}{\left(2x^{2}\right)}=3\) - step3: Convert the logarithm into exponential form: \(2x^{2}=4^{3}\) - step4: Evaluate the power: \(2x^{2}=64\) - step5: Divide both sides: \(\frac{2x^{2}}{2}=\frac{64}{2}\) - step6: Divide the numbers: \(x^{2}=32\) - step7: Simplify the expression: \(x=\pm \sqrt{32}\) - step8: Simplify: \(x=\pm 4\sqrt{2}\) - step9: Separate into possible cases: \(\begin{align}&x=4\sqrt{2}\\&x=-4\sqrt{2}\end{align}\) - step10: Check if the solution is in the defined range: \(\begin{align}&x=4\sqrt{2}\\&x=-4\sqrt{2}\end{align},x>0\) - step11: Find the intersection: \(x=4\sqrt{2}\) The solution to the equation \( \log_{4}x+\log_{4}2x=3 \) is \( x=4\sqrt{2} \).