$$3^{x-1}+3^{x-2}+3^{x-3}=3159$$
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.
On
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 $$3^x = 3^8.$$
You could use $3^{x-1} = 9 \cdot 3^{x-3}$ and $3^{x-2} = 3 \cdot 3^{x-3}$ to get
$$3^{x-1}+3^{x-2}+3^{x-3} = 3^{x-3} (9 + 3 + 1) = 3^{x-3} \cdot 13 = 3159$$
Dividing both sides by $13$ gives
$$3^{x-3} = 243 = 3^5$$
So then $x - 3 = 5$ or equivalently $x = 8$.