Solving a logarithmic equation with a mix of polynomials and logarithms How can I solve the equation $\dfrac x9 = \log_2x$ ? 
 A: If you plot the graphs of $y=\frac{x}{9}$ and $y=\log_2 x$, you will see that there are two solutions: $1 < x_1 < 2$ and $5 < x < 6$. You can improve the accuracy by using, e.g., Newton's method.
A: I was having trouble with a similar equation, $x^2+ln(x) = y$, and I found the Lambert W function, that is useful to solve equations in which the unknown appears both outside and inside an exponential or a logarithm.
Basically $W=f(x)^{-1}$, where $f(x) = xe^x$, so $W(xe^x) = x$.
We have $\frac{x}9=log_2(x)$ that can be written as $\frac{x}9=\frac{ln(x)}{ln(2)}$ so we have e as the base, and then we can rewrite it as $\frac{ln(2)}{9}=\frac{ln(x)}{x}$, then we solve.
$\require{enclose}$
$$\frac{ln(2)}{9}=\frac{ln(x)}{x}$$
$$\frac{ln(2)}{9}=\frac{ln(x)}{e^{ln(x)}}$$
$$\frac{ln(2)}{9}={ln(x)}{e^{-ln(x)}}$$
$$-\frac{ln(2)}{9}=-{ln(x)}{e^{-ln(x)}}$$
$$W(-\frac{ln(2)}{9})=W(-{ln(x)}{e^{-ln(x)}})$$
$$W(-\frac{ln(2)}{9})=-ln(x)$$
$$\bbox[5px,border:2px solid black]{
x_1=e^{-W(-\frac{ln(2)}{9})} ,\; x_2=e^{-W_{-1}(-\frac{ln(2)}{9})}
}$$
You can check the result in WolframAlpha to see the exact values. Hope this was useful to someone looking for ways to solve this type of equations.
