Solving $y = x + \log_2x$ for $x$ Can you solve for $x$ in $y = x + \log_2x$ and show how you do it? Looking at the graph it seems like the inverse of this function can exist and be defined on the entire domain. I have no idea what to do with the $\log_2$, exponentiating both side doesn't seem to lead anywhere for me. Also, I found through guesswork that $y\ =\ x-2^{-x}$ seem to be a 45 degree symmetry to the solution.
 A: \begin{align}
y&=x+\log_2{(x)}\\
y&=\log_2{(2^x)}+\log_2{(x)}\\
y&=\log_2{(x2^x)}\\
2^y&=x2^x\\
2^y&=xe^{\ln{(2)}x}\\
\ln{(2)}2^y&=\ln{(2)}xe^{\ln{(2)}x}\\
W_k(\ln{(2)}2^y)&=\ln{(2)}x\\
\therefore x&=\frac{W_k(\ln{(2)}2^y)}{\ln{(2)}}\qquad k\in\mathbb{Z}\\
\end{align}
where $W_k$ denotes the $k$th branch of the Lambert W Function.
A: \begin{align*}
y &= x + \log_2 x  \text{,}  \\
2^y &= 2^{x + \log_2 x}  \\
    &= 2^x \cdot 2^{\log_2 x}  \\
    &= (\mathrm{e}^{\ln 2})^x \cdot x  \text{, and}  \\
2^y \ln 2 &= \mathrm{e}^{x \ln 2} \cdot x \ln 2  \text{, so }  \\
x \ln 2 &= W_k(2^y \ln 2)  \text{ and finally}  \\ x &= \frac{W_k(2^y \ln 2)}{\ln 2}  \text{,}
\end{align*}
where $W_k$ is the/a Lambert $W$ function.  Since you are likely thinking of $x$ as a real number, you probably intend the real logarithm to the base $2$, so $x > 0$ and so you are only interested in the $k = 0$ branch of the $W$ function.
A: Raise the equation to the power of $2$
\begin{eqnarray*}
2^y= x2^x = xe^{x \ln(2)}.
\end{eqnarray*}
Now multiply by $\ln(2) $
\begin{eqnarray*}
\ln(2) 2^{y} = (\color{red}{x \ln(2)})e^{\color{red}{x \ln(2)}}.
\end{eqnarray*}
Now recall the Lambert $W$ function is defined by $we^w=z$ gives $w=W(z)$. So we have
\begin{eqnarray*}
x \ln(2) =W(\ln(2) 2^{y} ) \\
x  = \frac{1}{\ln(2)} W(\ln(2) 2^{y} ). \\
\end{eqnarray*}
