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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.



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.

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It's hard to say whether you are making a mistake along the way if you don't show us what you've done...(-: – Arturo Magidin Aug 28 '11 at 19:14
As a general rule, I would suggest that if you keep numbers as small as possible, you are less likely to make arithmetical errors. I would myself start by noting that $3159$ is divisible by $9$ (a positive integer is whenever the sum of its decimal digits is). Hence $3^{x-3} +3^{x-4}+3^{x-5} = 351$. The right side is again divisible by $9$, so $3^{x-5} + 3^{x-6}+ 3^{x-7} = 39$. Then $3^{x-6}+3^{x-7} + 3^{x-8} = 13.$ – Geoff Robinson Aug 28 '11 at 21:57
up vote 6 down vote accepted

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 $x = 8$.

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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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@Thijs: Absolutely true. – Arturo Magidin Aug 28 '11 at 19:23
Ah, again I'm in a spot where I have to choose between 2 correct answers. Thank you both so much for answering. – BeatShot Aug 28 '11 at 19:27

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