How to find the number of solutions of equation $x^n - a^x = 0$?

Question

I have to find the number of solutions of the equation $x^4 - 5^x = 0$

Since it is only asked to find the number of solutions and not the exact solution, what is the best way to approach such equations (of the form $x^n - a^x = 0$)?

Thanks a lot!

Answer 1

The easiest way is to sketch or plot the curve, like so. But you can also reason it out. You know that $5^x > 0 \forall (x \in \mathbb{R})$ with $\lim_{x \to -\infty} 5^x = 0^+$ and $5^x$ is strictly increasing (i.e. $(5^x)' = \ln 5\cdot 5^x>0 \forall (x \in \mathbb{R})$) with $5^0=1$.

Now $x^4$ is an even function that passes through $(-1,1), (0,0)$ and $(1,1)$. Also $\lim_{x \to -\infty} x^4 = \infty$

Let's define $g(x) = x^4 - 5^x$. Now $g(0) = -1$. You also know that $g(x) > 0$ for sufficiently negative values of $x$ because of the behaviour of both the curves as $x \to -\infty$. Of course, $g(x)$ is continuous. The intermediate value theorem therefore guarantees the existence of a value $X$ (a negative real number) such that $g(X) = 0$. You can now prove uniqueness by considering the rate of change of gradient of each curve for $x \geq 0$. Therefore, there is exactly one solution to the equation.

Answer 2

It may be a non rigorous method but for numbers like the ones in your example, it's easy to visualize the geometry of each curve. If you want $x^4-5^x=0 \iff x^4=5^x$ you just have to imagine roughly on how many points these two curves meet. There will probably be a point where they meet in the low negatives, but no point in the positives, as $5^x$ grows faster than $x^4$ for $x>0$. For $x^n-a^x=0$ it depends it's better to separate it into cases where $n$ is odd/even and $a>1$ or $a<1$ then study their growth rates. Here is the plot of your two curves.