\documentclass[a4paper]{article}
\usepackage{amsmath}
\begin{document}
\title{Solving $3^x-2^x=65$}
\author{Amit Yaron}
\date{Jan 21, 2025}
\maketitle
\section{"Most People Don't Even Know Where to Start"}
Another one from YouTube with "Most people don't event know where to start"
in the title. Really? Start by finding where solutions lie, and the number of
them. See if guess-and-check is enough or if approximation is needed.
\section{Where to Find Solutions}
Let us check what values $x$ can have: zero? Positive? Negative? We can 
determine it using the following inequality:
\begin{align*}
& 3^x-2^x=65\Rightarrow\\
\Rightarrow & 3^x-2^x>0\Rightarrow\\
\Rightarrow & 3^x>\underbrace{2^x}_{>0}\Rightarrow\\
\Rightarrow & \frac{3^x}{2^x}>1\Rightarrow\\
\Rightarrow & \big(\frac32\big)^x>1\Rightarrow\\
\Rightarrow & x>0
\end{align*}
\section{Increasing or Decreasing?}
Let us use tha multiplication rule:
\[ (ab)^x=a^xb^x \]
And 
\[
  3=2\cdot\frac32
\]
Thus,
\begin{align*}
  3^x-2^x&=\big(2\cdot\frac32\big)^x-2^x\\
         &=2^x\big(\frac32\big)^x-2^x\\
         &=2^x\bigg(\big(\frac32\big)^x-1\bigg)
\end{align*}
Now, $2^x$ is increasing, and so is $\big(\frac32\big)^x-1$\\
Thus, their product is increasing, and that product is $3^x-2^x$.\\
It means that in the equation
\[
  3^x-2^x=65
\]
The left-hand side is increasing, while the right-hand side is constant.\\
Thus, there is only one real solution.
\section{Is $x$ an Integer?}
One way to find it is to try looping over positive integers until we find
a place to stop. But, we can also try other things>
$3^x-2^x$ is a difference. Is it a difference of two squares? Let us try
using the two identities:
\[ a^2-b^2=(a+b)(a-b) \]
and
\[ {(a^b)}^c=a^{bc} \]
So,
\begin{align*}
 3^x-2^x&=3^{2(\frac{x}2)}-2^{2(\frac{x}2}\\
        &=(3^{\frac{x}2}-2^{\frac{x}2})(3^{\frac{x}2}+2^{\frac{x}2})
\end{align*}
and 65 can be factored over the natural numbers in two ways:
\[ 65=13\cdot5 \]
and
\[ 65=65\cdot1 \]
Let us try the first:
\[
\begin{cases}
  3^{\frac{x}2}+2^{\frac{x}2}=13\\
  3^{\frac{x}2}-2^{\frac{x}2}=5
\end{cases}
\]
By adding both equations we get that:
\begin{align*}
& 2\cdot 3^{\frac{x}2}=18\Rightarrow\\
\Rightarrow &3^{\frac{x}2}=9=3^2\Rightarrow\\
\Rightarrow &\frac{x}2=2\Rightarrow\\
\Rightarrow &x=4
\end{align*}
Let us test the solution:
\[ 3^x-2^x=3^4-2^4=81-16=65 \]
Bingo! And it is the only solution.
\end{document}
