$\frac{8^x-2^x}{6^x-3^x}=2$, Solve for $x$

Question

So I was looking through the homepage of Youtube to see if there were any math equations that I thought that I might be able to solve when I came across this video by LKLogic asking to solve for $x$ in $$\color{black}{\dfrac{8^x-2^x}{6^x-3^x}=2}$$ which I thought that I might be able to do. Here is my attempt for solving for $x$ in $\dfrac{8^x-2^x}{6^x-3^x}=2$:

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$$\dfrac{8^x-2^x}{6^x-3^x}=2$$I then realized (after my first attempt taking up nearly a full page) that $8-2=6$, $6-3=3$, $6/3=2$, $\therefore x=1$, therefore I somehow managed to waste time without realizing that $x=1$ right away. In the following spoiler (for anyone who is interested), here is my first attempt in full detail:

$$\dfrac{8^x-2^x}{6^x-3^x}=2$$$$\dfrac{2^{3x}-2^x}{2^x3^x-3^x}=2$$$$\dfrac{2^x(2^{2x}-1)}{3^x(2^x-1)}=2$$$$2^x(2^x2^x-1)=2\cdot3^x(2^x-1)$$$$2^{x-1}=\dfrac{3^x(2^x-1)}{2^{2x}-1}$$I'm not even going to show you the rest of the first attempt, just know that I somehow got $2^{2x-2}-0.25=3^x$ which doesn't even equal each other at $1$ (it intersects somewhere around $4.8$ for anyone who wants to know)

My question

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Is the solution that I have arrived at correct, or what could be an alternate way that I could formulate my proof?

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To clarify

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1. The only reason I'm not showing the entirety of my first attempt is because I really just want to spare you the pain that is my (obviously) incorrect first attempt at solving the math olympiad equation and in all honesty, I don't even know what I was thinking there :\

Answer 1

Suppose $x>0$. The equality can be written as $$ \left(\frac{2^{x}+1^{x}}{2}\right)^{1/x}=\frac{3}{2} $$ for $x\neq 0$. It holds for $x=1$ by direct substitution.

By the generalized mean inequality the function $\left(\frac{2^{x}+1^{x}}{2}\right)^{1/x}$ is strictly increasing. Hence the previous equality holds only for $x=1$.

Suppose $x<0$. Put $y=-x>0$. The equality can be written as $$ \left(\frac{\left(1/2\right)^{y}+1^{y}}{2}\right)^{1/y}=\frac{2}{3} $$ for $y\neq 0$. When $y\rightarrow0$ we have that the left side of this equation tends to the geometric mean of $1$ and $1/2$ that is $\sqrt{1/2} > 2/3$. Then there is no solution in this case too since, as before, the function on the left side of the equation is striclty increasing.

Therefore the solution $x=1$ is unique in the original equation.

Answer 2

First, note $x\neq0$

$$2=\dfrac{8^x-2^x}{6^x-3^x}=\frac{2^x(2^{2x}-1)}{3^x(2^2-1)}=\frac{2^x(2^x+1)}{3^x}=\frac{4^x+2^x}{3^x}$$

We get:

$$2=\left(\frac{4}{3}\right)^x+\left(\frac{2}{3}\right)^x=f(x)$$

Since

$$f''(x)=\left(\frac{4}{3}\right)^x\ln^2\left(\frac{4}{3}\right)+\left(\frac{2}{3}\right)^x\ln^2\left(\frac{2}{3}\right)>0, ~~\forall x\in \mathbb{R}$$

it means function $f(x)$ is a strictly convex function on $\mathbb{R}$, hence it has at most two roots for $f(x)=2$. Namely, $f(0)=f(1)=2$. But $x\neq0$, so $x=1$ is the only solution.

Answer 3

$$2 \cdot (6^x-3^x) = 8^x-2^x \\ 2^1 \cdot 3^x \cdot (2^x-1) = 2^x(4^x-1) $$ Because on the lhs there is only 2 to the power of 1 as factor, the same must occur on the rhs, and thus ...

• the only solution (if there is one at all) has $x=1$

With this $$2 \cdot (6^1-3^1) = 8^1-2^1 \\ 2 \cdot 3 = 6 $$ and

• the full equation holds for $x=1$ (which is the only solution).