Calculus – How to Solve the Equation x+1=5e^{4x}

Question

How to solve $x+1=5e^{4x}$

In general, I know to take ln() of both sides to bring down the exponent for e, but the left side is also a variable.

Answer 1

Rewrite the equation like this

$$x+1 = 5e^{4x} = 5e^{4(x+1)-4}$$

$$(x+1)e^{-4(x+1)} = 5e^{-4}$$

$$-4(x+1)e^{-4(x+1)} = -20e^{-4}$$

which has two solutions. There is the standard Lambert-W function

$$W\left(-4(x+1)e^{-4(x+1)}\right) = W\left(-20e^{-4}\right)$$

$$-4(x+1) = W\left(-20e^{-4}\right)$$

$$x = -\frac{1}{4}W\left(-20e^{-4}\right) - 1 \approx -0.7724_\cdots$$

and the other real branch of the Lambert-W, the $-1$ branch

$$W_{-1}\left(-4(x+1)e^{-4(x+1)}\right) = W_{-1}\left(-20e^{-4}\right)$$

$$-4(x+1) = W_{-1}\left(-20e^{-4}\right)$$

$$x = -\frac{1}{4}W_{-1}\left(-20e^{-4}\right) - 1 \approx -0.7262_\cdots$$

Graphically, the two real branches occur whenever you are taking the Lambert-W of a number which lies in the interval $\left(-\frac{1}{e},0\right)$. As seen below, there are two real solutions for such numbers - one on either side of the local minimum of $-\frac{1}{e}$ that occurs at $x=-1$ for the function $f(x) = xe^x$