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 $\dfrac{8^x-2^x}{6^x-3^x}=2$:
$$\dfrac{8^x-2^x}{6^x-3^x}=2$$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
Is the solution that I have arrived at correct, or what could be an alternate way that I could formulate my proof?
To clarify
- 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 :\
Best Answer
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.