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, $x=2$

Here's my resolution :)
$$
2^\frac{2x-1}{x-1} + 2^\frac{3x-2}{x-1} = 24 \\
\left(2^\frac{1}{x-1}\right)^{2x} \ \left(2^\frac{1}{x-1}\right)^{-1} + \left(2^\frac{1}{x-1}\right)^{3x} \ \left(2^\frac{1}{x-1}\right)^{-2} = 24  \\
2^\frac{2x-1}{x-1} \ \left[ 1 + 2^\frac{x-1}{x-1}  \right] = 24\\
2^\frac{2x-1}{x-1} \cdot 3  = 24\\
\log_2 \ 2^\frac{2x-1}{x-1} = 8 \\
\frac{2x-1}{x-1} \cdot \log_22= \log_2 8\\
2x-1=3(x-1)\\
x=2
$$
