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I was helping a high school student with their homework and I ran across the following problem: solve $$3^x+3^{(3x+1)}=108.$$ I was unable to find any "elementary way" to do this (by which I mean something a high school student would be comfortable with like manipulating the equation until the bases are the same and then equating the power, or using logarithms to undo exponents).

Can anyone solve this with $9^{th}$ or $10^{th}$ grade level math?

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    $\begingroup$ Replace $y=3^x$. You don't get a nice cubic, but it boils down to solving $y+3y^3=108$. Are you sure that this is the question? Cubics are always solvable by different methods, but in this case solutions are not pretty $\endgroup$ – chubakueno Aug 8 '14 at 0:18
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    $\begingroup$ Are you sure the question wasn't $3^x + 3^{x+1} = 108$ ? $\endgroup$ – DanielV Aug 8 '14 at 0:24
  • $\begingroup$ That is the method i used (i basically walked the student through cardano's formula). But the student lost interest and the teacher explicitly stated "show all steps" and no calculator. I too was thinking there was a typo but that was not comforting to the student. $\endgroup$ – illysial Aug 8 '14 at 0:24
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    $\begingroup$ Definitely a typo! Or they want the student to use a calculator to solve... $\endgroup$ – ClassicStyle Aug 8 '14 at 0:35
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    $\begingroup$ @TylerHG certainly no calculators allowed.. And I'm glad others agree me..I will show this to the student and hopefully he will believe me! $\endgroup$ – illysial Aug 8 '14 at 0:36
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Follow chubakueno's suggestion of letting $y=3^x$, giving us

$$3y^3+y-108=0$$

The only hope to factor this is for the equation to have a rational root. According to the rational root theorem, such a root would have to be of the form $\frac pq$ such that $p$ is a factor of $108$ and $q$ is a factor of $3$. It's also relatively easy to see that there should be one real root between $3$ and $4$. Since our root is not an integer, we must test $q=3$. Since neither $10$ nor $11$ are factors of $108$, this root cannot be rational. High school algebra won't be able to solve this one.

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  • $\begingroup$ You need to assume $\,p,q\,$ coprime for RRT to apply, i.e. that $\,p/q\,$ is in lowest terms. $\endgroup$ – Bill Dubuque Aug 8 '14 at 0:58
  • $\begingroup$ @BillDubuque If $p$ and $q$ aren't coprime, the numerator and denominator of the reduced fraction should still be factors of the proper coefficients. $\endgroup$ – Mike Aug 8 '14 at 1:09
  • $\begingroup$ RRT requires coprimality, e.g. $\,2/6\,$ is a root of $\,3x\!-\!1\,$ but $\,6\nmid 3,\ 2\nmid 1.\ $ Yes, wlog you may assume the fraction is in lowest terms. But you must say that to apply RRT. $\endgroup$ – Bill Dubuque Aug 8 '14 at 1:16
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    $\begingroup$ @BillDubuque I think you're just nitpicking. Is it so wrong that I find the root $1/3$ and miss the root $2/6$? They're equal. If you try all such $p$ and $q$ as laid out here, you will find all roots. It may be possible to write them in different forms, but who cares? The point is to set up a finite search space to find all roots. $\endgroup$ – Mike Aug 8 '14 at 1:46
  • $\begingroup$ This oversight can lead to subtle, serious errors in less trivial contexts. That's why when I teach RRT I always emphasize proper application. I recall one student who wasted a few days searching for a subtle error in his thesis that essentially boiled down to this oversight. But I can certainly understand why it may seem like nitpicking. $\endgroup$ – Bill Dubuque Aug 8 '14 at 1:59

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