Number of solutions of $e^x=x^3$

How can I analytically obtain the number of solutions of the equation $$e^x=x^3$$?

I know Lambert's $$W$$ function, but even using that, we need to know the value of $$W(-1/3)$$ which I think we cannot calculate without a calculator/graph/etc.

I do not need the exact solutions of the equation, I just need to know the number of solutions it has. Is there any good approximation?

I tried using a method of comparing slopes of the two functions on the L.H.S and R.H.S , but it was turning out to be too lengthy.

• Are you asking for the specific solutions for this problem or a general method? – Ryan Shesler Apr 28 at 13:34
• @DavidMitra A typo by me. I should have said Clearly there are no negative solutions. You should also know $0^3 \lt e^0$ and $3^3 \gt e^3$ and $16^3 = 2^{12} \lt e^{16}$, so by continuity there should be at least two solutions. Now look at derivatives to show there are exactly two – Henry Apr 28 at 13:44

Since $$e^x>0$$, the equation is equivalent to $$x=3\log x$$. Consider $$f(x)=x-3\log x$$ (for $$x>0$$. The limit at $$0$$ and $$\infty$$ are both $$\infty$$.

The derivative is $$f'(x)=1-3/x=(x-3)/x$$. So the function has an absolute minimum at $$x=3$$ and $$f(3)=3-3\log 3=3(1-\log3)$$. Thus the equation has exactly two solutions.

In order to approximate them you can use any numerical method. One is $$\approx1.8571$$.

Suppose you have $$e^x=x^2$$; then the function to study is $$f(x)=x-2\log\lvert x\rvert$$. The limits are $$\lim_{x\to-\infty}f(x)=-\infty,\quad \lim_{x\to0^-}f(x)=\infty,\quad \lim_{x\to0^+}f(x)=\infty,\quad \lim_{x\to\infty}f(x)=\infty$$ and $$f'(x)=1-\frac{2}{x}=\frac{x-2}{x}$$ The function is increasing over $$(-\infty,0)$$ and $$[2,\infty)$$; decreasing over $$(0,2]$$.

Since $$f(2)=2-2\log2<0$$, we have one solution in $$(-\infty,0)$$ and two solutions in $$(0,\infty)$$.

• Reading your argument I have another idea. $3\log(x)$ is a concave function. At $x=3$ the staight line $y=x-3+3\log(3)$ is a tangent. Since $3\log(3)-3>0$, the straight line $y=x$ is a secant for $y=3\log(x).$ By concavity, if cuts this curve at exactly two points. – szw1710 Apr 28 at 13:41
• @szw1710 That's an alternative way. On the other hand, knowing that $f(1)=1$ and $f(2)=2-3\log2=\log(e^2/9)<0$, we know that one solution is in $(1,2)$. Also $f(4)=2(2-3\log2)<0$ and $f(5)>0$, so the other solution is in $(4,5)$. – egreg Apr 28 at 13:45
• If instead of x³ ,I had x², now, going by your method how can we get that 1 negative solution? – user226375 Apr 28 at 13:53
• @user226375 You just solve two equations: $x=2\log x$ and $x=2\log(-x)$. – egreg Apr 28 at 13:58
• @user226375 I added the sketch for the case $e^x=x^2$. – egreg Apr 28 at 14:08

I understand that you are not looking for an exact solution. Nevertheless it is for interest to show the analytical solution : $$e^x=x^3$$ $$e^{x/3}=x$$ $$xe^{-x/3}=1$$ $$\frac{-x}{3}e^{-x/3}=-\frac13$$ According to the definition of the Lambert W function : $$\quad \frac{-x}{3}=W\left(-\frac13 \right)$$

The Lambert W function is multivalued. In the real domain the two branches are noted $$W_0$$ and $$W_{-1}$$. So they are two real solutions : $$x=-3W_0\left(-\frac13 \right)\simeq 1.857183$$ $$x=-3W_{-1}\left(-\frac13 \right)\simeq 4.536403$$

Approximates :

$$W_0(x)\simeq-\sum_{k=1}^n \frac{(-1)^k k^{k-1}}{k!}x^k$$ The series is very slowly convergent. For enough accuracy $$n$$ must be large.

$$W_{-1}(x)\simeq -\theta-\ln(\theta+\ln(\theta+\ln(\theta+...)))$$ with $$\theta=-\ln(-x)$$ .

The number of terms must be large for enough accuracy.

The above recursive formulas are not recommended (except with computer). Better use numerical calculus with iterative method such as Newton-Raphson.

Not using Lambert function, consider that you look for the zero's of $$f(x)=e^x-x^3$$But easier would be $$g(x)=x-3\log(x)$$for which $$g'(x)=1-\frac 3x \qquad \text{and} \qquad g''(x)=\frac 3{x^2} > 0 \qquad \forall x$$ The first derivative cancels when $$x_*=3 \implies g(3)=3-3\log(3) <0$$ So, two roots.

To approximate them, use a Taylor series around $$x=3$$ and get $$g(x)=(3-3 \log (3))+\frac{1}{6} (x-3)^2+O\left((x-3)^3\right)$$ Ignoring the higher order terms, the solutions are given by $$x_\pm=3\pm 3 \sqrt{2 (\log (3)-1)}$$ that is to say $$1.6677$$ and $$4.3323$$; not too bad for a first approximation.

If you want to polish the roots, using Newton method we should get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.667700786 \\ 1 & 1.834637542 \\ 2 & 1.856830114 \\ 3 & 1.857183772 \\ 4 & 1.857183860 \end{array} \right)$$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 4.332299214 \\ 1 & 4.546901400 \\ 2 & 4.536427193 \\ 3 & 4.536403655 \end{array} \right)$$

Edit

Sooner or later, you will learn that, better than with Taylor series, functions can be locally approximated using Padé approximants. Using the $$[2,2]$$ one (not too complex to build), we shoulf get $$g(x)=\frac{(3-3 \log (3))-\frac{2(\log (3)-1)}{3} t +\frac{(8+\log (3))}{54} t^2 } {1+\frac{2 }{9}t-\frac{1}{162}t^2}$$ where $$t=(x-3)$$.

Solving the quadratic in numerator will lead to $$t=\frac{9 \left(2 \log (3)-2\pm\sqrt{6 \left(-2+\log ^2(3)+\log (3)\right)}\right)}{8+\log (3)}$$ given then, as estimates , $$x_1=1.85574$$ and $$x_2=4.53443$$.