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This is the equation I have:

$$2^{2x} + 9e^{-2x} = 6$$

I want to solve for x using the substitution method.

I've turned it into

$$4^x+\frac{9}{e^{2x}} - 6 = 0$$

But I do not know what to substitute and how to solve it.

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Hint: Multiply each term by $e^{2x}$ and then, rewrite each term using only base $e$. –  NasuSama May 12 '13 at 14:00
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2 Answers

up vote 3 down vote accepted

Hints:

$$2^{2x}=e^{2x\log2}\implies 2^{2x} + 9e^{-2x} = 6\implies e^{2x(1+\log2)}-6e^{2x}+9=0\implies\ldots$$

Edit: Perhaps, as suggested in the comments, there's a typo in the OP, which misled me to commit a typo myself and a mistake: if the equation were

$$e^{2x}+9e^{-2x}=6\implies\left(e^{2x}\right)^2-6e^{2x}+9=0$$

and all is nice and dandy with the quadratic $\,t^2-6t+9=(t-3)^2\,$ ... As it stands now it looks like a mean, evil exponential equation of the form $\,t^{1+\log 2}-6t+9=0\,$ ...

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I don't understand the second implication.This one---> $2^{2x} + 9e^{-2x} = 6\implies e^{1+\log2}e^{2x}-6e^{2x}+9=0$ –  boywholived May 12 '13 at 15:10
    
@boywholived: Don't understand it either. Probably there is a typo in the original question. –  André Nicolas May 12 '13 at 15:54
    
I made a typo and a mistake, misled by thinking first that $\,2^{2x}\,$ was $\,e^{2x}\,$... I edited my answer. –  DonAntonio May 12 '13 at 16:27
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Yeah there is a typo in the original question.. It shouldn't be turned into an equation that requires the Lambert W-function to solve. But thanks for answering my question. This is very useful. –  Just a guy May 13 '13 at 6:24
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This based on the equation

$$ a^b = e^{b \log a}. $$

Apply this to the only exponential not with base $e$ in your problem:

$$ \begin{aligned} 2^{2x} + 9e^{-2x} &= 6\\ e^{2x \log 2} + 9e^{-2x} &= 6 \end{aligned} $$ Now, the only substitution for $x$ I see is $u = e^{2x}.$

$$ \begin{aligned} u^{\log 2} + \frac{9}{u} &= 6\\ u^{1 + \log 2} + 9 &=6u\\ u^{1 + \log 2} -6u + 9 &=0 \end{aligned} $$

I don't see any way to solve for $u$, and so for $x$, here. Something involving the Lambert $W$-function, perhaps.

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