Quadratic equation with different indices 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.
 A: There was a typo in the question. It should have been
$$729+3^{2x+1}=4\times3^{x+3}$$
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.
A: Hint:
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 $\,\color{red}{x=2}$.

A: HINT
Set $3^x=y$
$$729 + 3y^2= 4\cdot 9 y$$
$$3y^2-36y+729=0$$

$$y^2-12y+243=0$$

$\Delta <0 \implies$ no real solution 
