If $2^\frac{2x-1}{x-1}+2^\frac{3x-2}{x-1}=24$, find all values of $x$ that satisfy this As title suggests, the problem is as follows:

Given that $$2^\frac{2x-1}{x-1}+2^\frac{3x-2}{x-1}=24$$ find all values of $x$ that satisfy this.

This question was shared in an Instagram post a few months ago that I came across today. Examining it at first, it seems there are many ways to solve this. I'll show my own approach here, please let me know if there are any issues in my solution and please share your own solution too!
Here's my approach for the problem:
Let $a=2^\frac{2x-1}{x-1}$ and $b=2^\frac{3x-2}{x-1}$
We then get $a+b=24$.
Now notice that:
$$(3x-2)-(2x-1)=x-1$$
That gives us a motivation to perform division with $a$ and $b$ (as the denominator and numerator of the exponent will be equal, hence reducing the exponent) thus:
$$\frac{b}{a}=\frac{2^\frac{3x-2}{x-1}}{2^\frac{2x-1}{x-1}}$$
$$\frac{b}{a}=2^\frac{x-1}{x-1}=2$$
$$b=2a$$
Therefore:
$$2a+a=24$$
$$3a=24$$
$$a=8$$
$$2^\frac{2x-1}{x-1}=8$$
$$2^\frac{2x-1}{x-1}=2^3$$
$$\frac{2x-1}{x-1}=3$$
$$2x-1=3x-3$$
Thus, $x=2$
 A: Here's my resolution :)
$$
2^\frac{2x-1}{x-1} + 2^\frac{3x-2}{x-1} = 24 \\
\left(2^\frac{1}{x-1}\right)^{2x} \ \left(2^\frac{1}{x-1}\right)^{-1} + \left(2^\frac{1}{x-1}\right)^{3x} \ \left(2^\frac{1}{x-1}\right)^{-2} = 24  \\
2^\frac{2x-1}{x-1} \ \left[ 1 + 2^\frac{x-1}{x-1}  \right] = 24\\
2^\frac{2x-1}{x-1} \cdot 3  = 24\\
\log_2 \ 2^\frac{2x-1}{x-1} = 8 \\
\frac{2x-1}{x-1} \cdot \log_22= \log_2 8\\
2x-1=3(x-1)\\
x=2
$$
A: First of all, your solution looks great.
Here's a slightly different way to discover that $x=2$. Noticing the $x-1$ denominators in the exponents, I would start by making the substitution $u = x - 1$, so $x = u + 1$. Thus,
$$
\frac{2x-1}{x-1} = \frac{2u+1}{u} = 2 + \frac{1}{u} 
$$
and
$$
\frac{3x-2}{x-1} = \frac{3u+1}{u} = 3 + \frac{1}{u}. 
$$
Now the left-hand side of the equation becomes
$$
2^{2 + \frac{1}{u}} + 2^{3 + \frac{1}{u}}
= 2^{2 + \frac{1}{u}} \bigl( 1 + 2 \bigr)
= 3 \cdot 2^{2 + \frac{1}{u}}, 
$$
so the equation reduces to
$$
2^{2 + \frac{1}{u}} = 8.
$$
Since the exponential function $t \mapsto 2^t$ is one-to-one on the real numbers$^\dagger$ and $8 = 2^3$, it follows that
$$
2 + \frac{1}{u} = 3,
$$
hence
$$
\frac{1}{u} = 1. 
$$
From there, it's clear that $u = 1$, so $x = 2$ is the only solution (which you can verify by plugging in!)
Note: the detour into the variable $u$ is not necessary, but it makes the relationship between the two exponents more obvious.
Challenge: Looking at the graph, it appears that the expression on the left-hand side of the equation is always decreasing on its domain. If you know differential calculus, prove this! This is another way to show that your solution is unique. Also, what is the domain? What is the range?

$^\dagger$This is no longer true if we are solving the equation in the complex numbers, where there will be an infinite family of solutions, stemming from the fact that $\exp(z + 2\pi k i) = \exp z$ for all integers $k$.
