Solving $\frac{8^x-2^x}{6^x-3^x}=2$

Question

$$\dfrac{8^x-2^x}{6^x-3^x}=2$$ It is easy to see that in the domain of $\mathbb{R}\setminus\{0\}$, the solution is $x=1$. https://www.desmos.com/calculator/dsei8j2sdq. Desmos adds that the only one. And formally?

PS. I transformed the given equation to the form: $(4^x-2\cdot3^x+2^x=0$ and the ideas are over :(

Answer 1

Find $x\neq 0$ such that $$\frac 12(2^x + 4^x) = 3^x\quad \Leftrightarrow\quad \frac{2^x + 4^x}{2} = \left(\frac{2 + 4}{2}\right)^x.$$ Use Jensen's inequality for:

• $0< x < 1$: For fixed $x$, the map $f:y\mapsto y^x$ is strictly concave, therefore for any $0 < a \neq b$, there holds $$ \frac{f(a) + f(b)}2 < f\left(\frac{a+b}{2}\right).$$ Use $a=2$ and $b=4$, which shows $\frac 12(2^x + 4^x) < 3^x$ for $0<x<1$, and therefore no equality is possible.
• $x > 1$ or $x<0$: Analogous argument, but with $f$ being strictly convex.

Hence, the only solution on $\mathbb R \setminus\{0\}$ is $x = 1$.

Answer 2

Alternative answer (using high school maths):

Dividing the transformed equation by $2^x$ yields $2^x + 1 = 2\cdot 1.5^x$. Define $f(x):=2^x + 1$ and $g(x) = 2\cdot 1.5^x$, so your problem amounts to showing that there are no solutions to $f(x) = g(x)$ other than $x\in\{0, 1\}$. (Note that we may consider the transformed equation on the whole of $\mathbb R$, and thus $x=0$ becomes an additional solution.)

Taking the first derivative yields $f'(x) = \ln(2) 2^x$ and $g'(x) =2\ln(1.5) 1.5^x$, and we find that $f'(x) > g'(x) \,\Leftrightarrow \, x>0.5455\ldots=:\tilde x$.

Therefore, $f(x) - g(x)$ is strictly monotonically increasing for $x>\tilde x$, and strictly decreasing for $x < \tilde x$. This means that there can be at most one solution (denote it by $x_L$) to $f(x) - g(x) = 0$ for $x<\tilde x$, and at most one solution (denoted by $x_R$) for $x>\tilde x$. But we already know those solutions: It is easily verified that $x_L = 0$ and $x_R = 1$. Having found those solutions, our proof tells us that there cannot be any further solutions to the equation.

Comment: The original problem, and the transformed problem involving $4^x - 2\cdot 3^x + 2^x = 0$ are equivalent on $\mathbb R\setminus\{0\}$ only, and the 'additional' solution $x_L = 0$ we found above is not a solution of the original problem. But this does not affect the validity of the above proof: If $x=0$ and $x=1$ are the only solutions of the transformed equation on $\mathbb R$, then $x=1$ will be the only solution on $\mathbb R\setminus\{0\}$ for the original equation.