<s>[INST] Exponential equations: Why is this method wrong? While solving exponential equations with my Y10 students, they found that you can break down a number into powers with the same base and add one $a^0$ and just use exponents, but never more than one power of one. They tried this out of laziness to do variable change and based on the fact that $a^{f(a)}=a^{g(a)} \rightarrow f(a) = g(a)$. Below is the part of the solution that contains the incorrect step, but why does it work?
$$
3^{x+1}-3^{-x+2}=26
$$
$$
26 = 27 - 1 = 3^3 - 3^0
$$
$$
3^{x+1}-3^{-x+2}=3^3-3^0
$$
$$
!!\rightarrow(x+1)-(-x+2)=3-0 \leftarrow!!
$$
$$
2x-1=3 
$$
$$
x = \frac{4}{2}=2
$$
Why does this work?
 [/INST] It works because the equation is actually involving a variation of the hyperbolic sine function $\sinh(x)=\frac{e^x-e^{-x}}{2}$ which is known to be invertible.
If $a$ is a positive real number different from $1$, let $\sinh_a(x)=\frac{a^x-a^{-x}}{2}$ (I will call it the "hyperbolic sine of base $a$"). Your equation is equivalent to
$$
\begin{array}{lcl}
& & \frac{3^{x+1}-3^{-x+2}}{2\times 3^{3/2}} = \frac{3^3-3^0}{2\times 3^{3/2}} \\
& \iff & \frac{3^{x-1/2}-3^{-(x-1/2)}}{2} = \frac{3^{3/2}-3^{-3/2}}{2} \\
& \iff & \sinh_3\left(x-\frac{1}{2}\right) = \sinh_3\left(\frac{3}{2}\right) \\
\end{array}
$$
$\sinh_a$ is easily shown to be one-to-one (for example because $sh_a(a)=\sinh(x\ln (a))$ and $\sinh$ is one-to-one), so $x-\frac{1}{2}=\frac{3}{2}$ which means that $x=2$.
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