My maths teacher gave me a worksheet to work through as I was getting slightly bored in lessons. However, there was one question which I cannot do. The worksheet gives the answer, but you are supposed to show how you did it. Here is the question:

$729 + 3^{2x+1} = 4\times3^{x+2}$

The sheet said that the answer was $x=2$ or $x=3$ Here is my working - I 'solved' by completing the square, as you will see:
\begin{align} 3^{2x+1}-4\times3^{x+2}+729& = 0 \\ 3(3^{2x})-4\times(3^2)(3^x)+729& = 0\\ 3(3^x)^2-36(3^x)+729& = 0\\ y &= 3^x\\ 3y^2-36y+729 & = 0\\ 3(y^2-12y+243) &= 0\\ (y-6)^2-36+243 &= 0\\ (y-6)^2+207 &= 0\\ (y-6)^2&=-207\\ y-6 &= \sqrt{207}i\\ y &= 6+\sqrt{207}i\\ 3^x &= 6+\sqrt{207}i\\ x &= \log_3(6+\sqrt{207}i)\\ \end{align} Clearly, this is not the answer as stated on the sheet. Have I answered this question wrong, or is the answer on the sheet wrong? Thank you in advance.

  • $\begingroup$ i got only complex Solutions of your equation! $\endgroup$ – Dr. Sonnhard Graubner Dec 13 '17 at 20:12
  • $\begingroup$ Clearly, neither $x = 2$ nor $x = 3$ is correct. I think your answer is correct but there is a typo in the second row, you wrote $3x$ but it should be $3^x$. $\endgroup$ – ArsenBerk Dec 13 '17 at 20:14
  • $\begingroup$ maybe there is a typo in your equation? $\endgroup$ – Dr. Sonnhard Graubner Dec 13 '17 at 20:14

There was a typo in the question. It should have been


Note the $+3$ in the exponent instead of $+2$.

(You are correct that the question as stated does not have integer solutions, in fact it has no real solutions).

There's actually some pretty mathematical content here though. We can rewrite $729=3^6$ and $4=1+3$ to make this expression: $$3^6+3^{2x+1}=3^{x+3}+3^{x+4}$$ Now the solutions $x=2$ and $x=3$ boil down to adding the same powers of $3$ on both sides. The completing the square method you outlined in your work will show that these are the only solutions.

  • $\begingroup$ could you please explain to me why $4\times3^x = 3^{x+3} + 3^{x+4}$? I am only a GCSE student, who is doing A-Level work, so I don't know how some of these concepts work yet $\endgroup$ – AJ123 Dec 13 '17 at 21:12
  • $\begingroup$ @AJ123 We can rewrite $4$ as $1+3$ and then distribute: $$4\times3^{x+3}=(1+3)\times3^{x+3}=1\times3^{x+3}+3\times3^{x+3}=3^{x+3}+3^{x+4}.$$ That being said, you certainly don't need to do this to solve the problem, I just think it looks nicer like this (to solve the question, all you need to do is complete the square like you showed). $\endgroup$ – Peter Woolfitt Dec 13 '17 at 21:29


This seems to be a problem in arithmetic. Note that $729=3^6$, and rewrite the equation as $$3^{2x+1}+3^6=4\cdot 3^{x+3}.$$

Some details

If $x$ is a natural number this implies $4$ divides the left-hand side. Depending on the values of $x$, factor out $3^6$ or $3^{2x+1}$. $4$ must divide the other factor. Now

  • if $2x+1>6$, we factor out $3^6$, obtaining $$3^6(3^{2x-5}+1)=4\cdot 3^{x+3},$$ so necessarily $\;x+3=6$ and $3^{2x-5}+1=4\iff 2x-5=1$. The first equality yields $\color{red}{x=3}$ and so does the second.
  • if $2x+1<6$, factoring out $3^{2x+1}$, we get $$3^{2x+1}(3^{5-2x}+1)=4\cdot 3^{x+3},$$ so $\,2x+1=x+3\,$ and $\,5-2x=1$. Both equations yield $\,\color{red}{x=2}$.
  • $\begingroup$ @PeterWoolfitt: I didn't say there were solutions. I explained a sensible way to tackle this problem. $\endgroup$ – Bernard Dec 13 '17 at 20:30
  • $\begingroup$ @PeterWoolfitt: See how it can be used in my completed answer. $\endgroup$ – Bernard Dec 13 '17 at 21:08
  • $\begingroup$ In the first case, how do we know $3^6=3^{x+3}$ and $3^{2x-5}+1=4$? Couldn't we also have some mixing of factors? e.g. $3^5=3^{x+3}$ and $3^{2x-5}+1=12$. I guess this does boil down to a finite number of checks though. $\endgroup$ – Peter Woolfitt Dec 13 '17 at 21:33
  • $\begingroup$ Unique factorisation.There's nothing to check: one of the factors is odd, the other is even. You don't have a choice. $\endgroup$ – Bernard Dec 13 '17 at 22:15
  • $\begingroup$ I think all you know so far is that one of the factors looks like $3^n$ and the other looks like $4\times3^m$. It seems to take a separate (if easy) argument to say $3^t+1=4\times3^m$ iff $m=0$. Still, now in the updated form (and using the correct question instead of the original), this answer is now acceptable. $\endgroup$ – Peter Woolfitt Dec 13 '17 at 22:25


Set $3^x=y$

$$729 + 3y^2= 4\cdot 9 y$$



$\Delta <0 \implies$ no real solution

  • $\begingroup$ ops of course sorry, thanks! $\endgroup$ – gimusi Dec 13 '17 at 20:18
  • $\begingroup$ sorry for the first incorrect version I’m working by the mobile! $\endgroup$ – gimusi Dec 13 '17 at 20:28
  • $\begingroup$ If you were to read my question you would see that I have already done that $\endgroup$ – AJ123 Dec 13 '17 at 20:52
  • $\begingroup$ ops sorry from mobile it’s difficult to see thing properly, in any case I think it’s the correct way to proceed ! $\endgroup$ – gimusi Dec 13 '17 at 20:54

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