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I'm dealing with a problem here. I'm trying to solve this exponential equation but I cannot find the solution: $$3^{x-1} + 3^{x-2} + 3^{x-3} + 3^{x-4}\cdot3^{x-5} + 3^{x-6}=364$$

Can anyone please tell me what to do ?

Thank you!

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    $\begingroup$ write 364 in base 3 $\endgroup$ – Will Jagy Jul 19 '15 at 18:26
  • $\begingroup$ @Student, Is this right equation-this wasn't your original post. $\endgroup$ – John_dydx Jul 19 '15 at 18:39
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    $\begingroup$ @john Somone edited it but now it is the right one. $\endgroup$ – Student Jul 19 '15 at 18:42
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    $\begingroup$ And you never thought about mentioning the fact that, so to speak, one plus is not a plus? $\endgroup$ – Did Jul 19 '15 at 18:50
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Multiplying the original equation by $3$:

$$3^x+3^{x-1}+3^{x-2}+3^{x-3}+3^{x-4}+3^{x-5}=1092$$

Now subtracting the original equation from this:

$$3^x-3^{x-6}=728$$ $$3^x-\frac{3^x}{729}=\frac{728}{729}3^x=728$$ $$3^x=729,x=6$$

Update: Now that the multiplication has been brought to my attention, it appears John's solution is best. I wish I could see some sort of algebra trick to employ to at least help factor anything, but I don't. It's just an ugly quadratic probably best suited to the quadratic formula.

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  • $\begingroup$ There is a subtlety overlooked. The OP wrote $3^{x-4}\cdot3^{x-5}$, not $3^{x-4}+3^{x-5}$ $\endgroup$ – Mark Viola Jul 19 '15 at 19:37
  • $\begingroup$ @Dr.MV Hmm... That does complicate things. Honestly, it was hard to look at before LaTex was added in. :) $\endgroup$ – Mike Jul 19 '15 at 20:23
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This equation can be re-written as:

$$\frac{3^x}{3}+ \frac{3^x}{9} + \frac{3^x}{27}+ \left(\frac{3^x}{81}\times \frac{3^x}{243}\right) + \frac{3^x}{729} = 364$$

let $3^x = a$ and multiply through by 19683.

$$ a^2 + 9504a -7164612 = 0$$

$$(a-702)(a+10206)= 0$$

$$a = 702 \ \text{or}\ a = -10206$$

Hence: $$3^x = 702$$

Taking $\ln$ of both sides:

$$x = \frac{\ln702}{\ln3} $$

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  • $\begingroup$ $3^x\ne702$ $ $ $\endgroup$ – Did Jul 19 '15 at 18:42
  • $\begingroup$ @Did, Please check OP's question carefully, most people here have solved the wrong question. $\endgroup$ – John_dydx Jul 19 '15 at 18:46
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    $\begingroup$ Ah, the product instead of an addition... How one can ask this question without even mentioning the fact is truly bewildering--but indeed you seem to be right. Kudos for the care with which you read the questions. :-) $\endgroup$ – Did Jul 19 '15 at 18:49
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    $\begingroup$ @John. I suppose that you mde a typo. From what you wrote, the last line is supposed to be $x = \frac{\ln702}{\ln3}$. I let you fixing tour good solution. $\endgroup$ – Claude Leibovici Jul 20 '15 at 9:33
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    $\begingroup$ You are very welcome ! $\endgroup$ – Claude Leibovici Jul 20 '15 at 10:01
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Another approach from the one Will suggested is this: recognize that the left hand side is a geometric series, namely $$ 3^{x-6} (1 + 3 + 3^{2} + ... + 3^{5}) $$ whose sum is $$ 3^{x-6} \cdot \frac{1 - 3^{6}}{1 - 3}. $$ Now the number $x$ appears only once, and you can do some algebra to get to a solution.

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  • $\begingroup$ There is a subtlety overlooked. The OP wrote $3^{x-4}\cdot3^{x-5}$, not $3^{x-4}+3^{x-5}$ $\endgroup$ – Mark Viola Jul 19 '15 at 19:37
  • $\begingroup$ Well ... sort of. I answered the question that was there when I wrote my answer. I also suspect that this question was the one that was intended by the question's author, whether or not it was transcribed correctly, because in the version with the $+$ sign, it turns out that $x$ is an integer, while in the other, it's just a mess. $\endgroup$ – John Hughes Jul 20 '15 at 2:10
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Write $364 = 3\cdot 11^2+1 = 3\cdot (3^2+2)^2+3^0$. Play around with expanding the quantity $3\cdot (3^2+2)^2$ and converting all terms to a power of $3$. This is tantamount to writing $364$ into base-$3$ and will immediately tell you if there is an $x$ that satisfies your equation. If you do this and don't get a basis representation that looks like your equation, then no solution exists because representations are unique.

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$$ 364=3^x \sum_{j=1}^6 3^{-j} = 3^{x-1} \sum_{j=0}^5 3^{-j} = 3^{x-1} \frac{1-3^{-6}}{1-3^{-1}}=3^x\frac{364}{3^6} $$ so $x=6$

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  • $\begingroup$ There is a subtlety overlooked. The OP wrote $3^{x-4}\cdot3^{x-5}$, not $3^{x-4}+3^{x-5}$ $\endgroup$ – Mark Viola Jul 19 '15 at 19:38

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