Solving $4x = e^x$ without graphing and looking for intersection

If I want to solve the equation $4x = e^x$, is there a way to solve for $x$ without graphing and looking for intersection?

Solution 1:

Let $y = -x$. Then $$\begin{align*} 4x &= e^x \\ y e^y &= -\frac{1}{4} \\ y &= W_k\left( -\frac{1}{4} \right), k \in \mathbb{Z} \\ x &= -W_k\left( -\frac{1}{4} \right), k \in \mathbb{Z} \end{align*}$$ where $W_k$ is the $k$th branch of the Lambert W function. Taking $k = 0$ or $-1$ results in real values.

Solution 2:

@HenryW. give you the solution in term of lambert W function that are: $$ x_1=-W(-\frac{1}{4}) \approx 2.15329... \qquad x_2=-W_{-1}(-\frac{1}{4}) \approx 0.357403.. $$

I want suggest a solution that does not use the graph but also does not use the Lambert function, but gives the solution as limits of sequences that you can find from the given equation.

Write your equation as: $$ x=\frac{1}{4}\exp(x) $$ now, replacing $x$ in RHS we have: $$ x=\frac{1}{4}\exp\left(\frac{1}{4}\exp(x)\right) $$ and replacing yet: $$ x=\frac{1}{4}\exp\left(\frac{1}{4}\exp\left(\frac{1}{4}\exp(x)\right)\right) $$ doing the same $n$ times you have a sequence of nested exponentials that converge to one solution of the equation (if it exists).

Using logarithms you can also write your equation as: $$ x=\log 4+\log x $$ and, with analogous substitutions we can find a sequence of nested logarithms:

$$ x=\log4+\log\left(\log 4+\log\left(\log 4 +\log x \right) \right) $$

and this also converge to a solution of the equation (if it exists).

A little numerical experiment with a spreadsheet shows that this works well , and the two sequences converge to the two solutions, as you can see in the figure.

Really I don't know if this works for all equations of this kind, but in this case it seems good.