Solving the exponential equation: $3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$

I have this exponential equation that I don't know how to solve: $3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ with $x \in \mathbb{R}$

I tried to factor out a term, but it does not help. Also, I noticed that: $2 \cdot 9^{x+1} = 2 \cdot 3^{2x+2}$
and tried to write the polynomial as a binomial square, without success.
I know I should solve it using logarithm, but I don't see how to continue.

Thank you,
rubik

EDIT: WolframAlpha factors it as: $(3 \cdot 2^x - 2 \cdot 3^x)(2^{x+2} - 3^{x+2}) = 0$ and then the solution is straightforward. Any hint about how to reach that?

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Oh yes sorry. I forgot to say it. I'll edit the question now, thank you. – rubik Feb 12 '12 at 9:17
@GeoffRobinson: Unfortunately I cannot use calculus tools because I haven't studied it yet. – rubik Feb 12 '12 at 9:18

$3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0 \Rightarrow 12 \cdot 2^{2x} - 35 \cdot 2^x \cdot 3^x + 18 \cdot 3^{2x} = 0 \Rightarrow$

$\Rightarrow 12 \cdot \left(\frac{2}{3}\right)^{2x}-35 \cdot \left(\frac{2}{3}\right)^{x}+18=0$

Now make substitution : $\left(\frac{2}{3}\right)^{x} = t$ , and solve quadratic equation .

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Thank you, I think this is the easiest solution. – rubik Feb 12 '12 at 9:24
+1, Very clever! – NoChance Feb 12 '12 at 11:05
@EmmadKareem: Yes, so I'll accept this answer! – rubik Feb 12 '12 at 11:13

The following substitution may have to work:

$$2^x=t; ~~3^x=s$$

Note that the equation simplifies to, $$12 t^2-35st+ 18s^2=0$$

This factorises to $$(3t-2s)(4t-9s)=0$$

Therefore,

$$3\cdot2^x=2 \cdot 3^x ~~\text{or}~~2^{x+2}=3^{x+2}$$ Since, $(2,3)=1$, we have that $\boxed{x=1~~ \text{or}~~-2}$

You can view that as a quadratic equation in $t$:

So, the solution will have to be, $$t=\dfrac{35s\pm\sqrt{(-35s)^2-4(12)(18s^2)}}{24}$$

This is a bit numerically taxing and honestly, I did not do it this way. Rather, I resorted to something that is equivalent to this. You need to write $18 \times 12$ as product of two numbers whose sum is $35$. Later, prefix a minus sign to these numbers and use them here.

With a little bit of playing around, with factorisations, you'll see they should be $27 \times 8$.

$$t=\dfrac{35s\pm\sqrt{(-35s)^2-4(12)(18s^2)}}{24} =\dfrac{35s\pm\sqrt{s^2[(-35)^2-4(12)(18)]}}{24} =\frac{35\pm 54}{24}s = \frac{2}{3}s\ or \frac{9}{4}s$$

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Thank you for the quick answer. Can you just elaborate on how to factorize that polynomial? See my EDIT. – rubik Feb 12 '12 at 9:24
Sorry, I has been unclear. I meant how to get from $12t^2 - 35st + 18s^2 = 0$ to $(3t - 2s)(4t - 9s)=0$. How could I ever see that factorization? – rubik Feb 12 '12 at 9:30
@rubik Is this convincing and clear? – user21436 Feb 12 '12 at 9:34
A slightly easier method (or I'm just used to it) would be to divide $12 t^2-35st+ 18s^2=0$ by $s^2$. Then let $\frac{t}s$ be $z$ and you get $12z^2-35z+18=0$, which is basically the same thing Kannappan added in answer, but I believe it's mentally easier to solve classical quadratic equation than dealing with a variable in quadratic solutions. Also Vieta's formulas could help in factorizing (that's where "You need to write 18×12 as product of two numbers whose sum is 35" comes from). – Lazar Ljubenović Feb 12 '12 at 9:48
@LazarLjubenović True. May be I could have been a bit more explicit as you say. Thank You. Do you think I should edit to add a mention of Viete? – user21436 Feb 12 '12 at 9:50

$$\begin{eqnarray} 0 &=& 3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} \\&=& 12 \cdot (2^x)^2 - 35 \cdot (2^x3^x) + 18 \cdot (3^x)^2 \\&=& 12 t^2 - 35 st + 18 s^2 \qquad\text{for}\qquad s=3^x,~t=2^x \\&=& (3t - 2s)(4t - 9s) \\ \implies&& s=\frac32t \quad\text{or}\quad s=\frac49t \\&& s=\left(\frac23\right)^\delta t \quad\text{for} \quad\delta=\frac{1\pm3}{2}=-1~\text{or}~2 \\&& 3^x=\left(\frac23\right)^\delta 2^x \\&& 1=\left(\frac23\right)^{x+\delta} \\&& 0=(x+\delta)\log\left(\frac23\right) \\&& x+\delta=0 \\&& x=-\delta=-\frac{1\pm3}{2}=\boxed{~1~\text{or}~-2~} \end{eqnarray}$$ To make the factorization in the fourth line, from the signs of the coefficients $12,-35,18$, we can see that we need to find a combination of the factors from a row on the left and a row on the right $$\begin{eqnarray} 1 \cdot 12 &\quad& 1 \cdot 18 \\ 2 \cdot 6 &\quad& 2 \cdot 9 \\ 3 \cdot 4 &\quad& 3 \cdot 6 \end{eqnarray}$$ so that a sum of the products (either of the first numbers and the second numbers, or of the first numbers with the second numbers) is $35$. So taking $3 \cdot 4$ for $12$ and $2 \cdot 9$ for $18$, we can obtain $2\cdot 4+3\cdot 9=8+27=35$ to get the factors above.

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Supplementing your answer with atleast a few words will enhance understanding. – user21436 Feb 12 '12 at 10:03

Set $a = 2^x, b = 3^x$, then $6^x = (2 \cdot 3)^x = 2^x \cdot 3^x = ab$ and $2^{2x+2} = 2^2 \cdot 2^{2x} = 4 \cdot 2^x \cdot 2^x = 4a^2$, $9^{x+1} = 9 \cdot 9^x = 9 \cdot 3^x \cdot 3^x = 9b^2$. So we get $$12a^2 - 35ab + 18b^2 = 0.$$ I think. What can you do with that?

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Thank you, this is the same as @Kannappan Sampath's answer. See my comment there. – rubik Feb 12 '12 at 9:25

from user21436 you have

$$t=\dfrac{35s\pm\sqrt{(-35s)^2-4(12)(18s^2)}}{24} =\dfrac{35s\pm\sqrt{s^2[(-35)^2-4(12)(18)]}}{24} =\frac{35\pm 54}{24}s = \frac{2}{3}s\ or \frac{9}{4}s$$

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This does not stand on its own as an answer; it could have been an edit. – Bookend Dec 22 '14 at 3:22
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. – gnometorule Dec 22 '14 at 3:33