How to find all solutions of $4^x-3^x=1$?

Question

I have problem with equation: $4^x-3^x=1$.

So at once we can notice that $x=1$ is a solution to our equation. But is it the only solution to this problem? How to show that there aren't any other solutions?

Answer 1

Let $f(x)=1-\left({3\over4}\right)^x-\left({1\over4}\right)^x$. The functions $(3/4)^x$ and $(1/4)^x$ are strictly decreasing functions of $x$, so the minus signs make $f(x)$ strictly increasing.

Answer 2

Hint: one way of showing that a function takes a value only once is to show that it is increasing.

Hint: For negative values of $x$ you need a different observation.

Answer 3

You can make a graph of the function $y=4^x-3^x-1=0$. It has a y-intercept at (0,-1) and for $x<0$ the curve stays under the y-axis and for $X>0$ the curve is only increasing.

Answer 4

Show that $4^x$ grows much faster than $3^x$.

Answer 5

Prove it by contradiction.

Let $f(x) = 4^x - 3^x - 1$. This function is smooth and as you have shown, $x = 1$ is a root. By Rolle's theorem, suppose it had two other roots $f(a) = f(b) = 0$, then there exists $c$ in between $a$ and $b$ (where $a < 1 < b$) such that $f'(c) = 0$, but notice for $ x \geq 0$

$$f'(x) = 4^x \ln 4 - 3^x \ln 3 \geq 3^x (\ln 4 - \ln 3) > 0.$$

and for $x < 0$, we have

$$f'(x) = 4^x \ln 4 - 3^x \ln 3 \leq \ln 4 (4^x - 3^x) < 0.$$

Hence you can conclude there is only one root.