<s>[INST] Is my outcome wrong? (Evaluating a logarithm) $$log\sqrt [ 4 ]{ x^2+y^2 } $$
$$log\sqrt { x+y } $$
$$logx^{ 1/2 }+log^{ 1/2 }$$
$$\frac { 1 }{ 2 }log (x+y)$$
The answer key saids:
$$\frac { 1 }{ 4 } log(x^2+y^2)$$
 [/INST] I'm afraid your answer is incorrect.
You should get this:
\begin{align*}
&\log \sqrt[4]{x^2+y^2}\\
&=\log(x^2+y^2)^{\frac{1}{4}}\\
&=\dfrac{1}{4}\log(x^2+y^2)
\end{align*}
You should note that $(a+b)^n\neq a^n+b^n$ in general, so $(a^2+b^2)^{\frac{1}{4}}\neq a^{\frac{1}{2}}+b^{\frac{1}{2}}$
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