Solving $y = x + \log_2x$ for $x$

Question

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.

Answer 1

\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.

Answer 2

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*}

Answer 3

\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.