<s>[INST] Solving an Exponential Equation $$3^{x-1}+3^{x-2}+3^{x-3}=3159$$
Another exponential equation I'm having a hard time with, the answer is given and equals to : $8$. I'm absolutely sure I'm making a wrong step somewhere along the way. Any help is appreciated.
EDIT:
$$3^{x-3}(\frac{1}{8}+\frac{1}{243}+1)=3159$$
This is where I'm now. If this is correct, it should pretty simple from here on. But I just don't seem to reach 8 as the final answer.
 [/INST] I would try factoring $3^x$ first, and isolating:
$$\begin{align*}
3^{x-1} + 3^{x-2} + 3^{x-3} &= 3159\\
3^x\left(3^{-1} + 3^{-2} + 3^{-3}\right) &= 3159\\
3^x &= \frac{3159}{3^{-1}+3^{-2}+3^{-3}}
\end{align*}$$
For an extra flourish,
$$\begin{align*}
\frac{3159}{3^{-1}+3^{-2}+3^{-3}} &= \frac{3159}{\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3}}\\
&= \frac{3159}{\quad\frac{9 + 3 + 1}{3^3}\quad}\\
&= \frac{3^3(3159)}{13}\\
&= \frac{3^3(3^5)(13)}{13}\\
&= 3^8.
\end{align*}$$
Now you see that your equation is equivalent to
$$3^x = 3^8.$$
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