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$$2^x+4^x+12=0$$

How exactly am I supposed to solve this? Am I supposed to get $x$ alone or solve it another way?

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    $\begingroup$ Hint: write $u = 2^x$ and find a quadratic in $u$. $\endgroup$
    – Simon S
    Commented Dec 11, 2014 at 3:36
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    $\begingroup$ Assuming that x is a real number, what do you know about any positive number raised to a real power (in this case $2^x $ and $4^x $) can it be negative? $\endgroup$
    – JMoravitz
    Commented Dec 11, 2014 at 3:39
  • $\begingroup$ In the future, please provide a more descriptive title; it makes it easier for users to help you. Also, users typically appreciate a more respectful tone than the imperative "solve this problem!" $\endgroup$ Commented Dec 11, 2014 at 4:16
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    $\begingroup$ If this is indeed precalculus . . . did you maybe transmute a - to a +? $\endgroup$
    – geometrian
    Commented Dec 11, 2014 at 5:20
  • $\begingroup$ What does "get $x$ alone" mean? $\endgroup$
    – JiK
    Commented Dec 11, 2014 at 10:05

2 Answers 2

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For any $x \in \mathbb{R}$, $2^x > 0$ and $4^x > 0$, therefore

$$ 2^x + 4^x + 12 > 0 + 0 + 12 = 12 > 0 $$

Therefore there is no real solution.

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    $\begingroup$ but no condition is given in the question stating $x\in R$ $\endgroup$ Commented Dec 11, 2014 at 3:43
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    $\begingroup$ @DheerajKumar The tag is "algebra precalculus", so what are the odd this is a complex variable exercise? $\endgroup$
    – Timbuc
    Commented Dec 11, 2014 at 3:45
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    $\begingroup$ You are right. But otherwise specified I think it is reasonabe to assume the convention $x$ denotes a member of $\mathbb{R}$ and $z$ denotes a member of $\mathbb{C}$. $\endgroup$
    – Empiricist
    Commented Dec 11, 2014 at 3:45
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let $u = 2^x$, then $4^x = (2^2)^x = 2^{2x} = 2^{2x} = (2^x)^2 = u^2$. Thus, $2^x + 4^x + 12 = 0$ becomes, $$u + u^2 + 12 = 0$$ Using the quadratic equation will solve $u$, which is really $2^x$. To solve for the $x$, just take $\log$ on both sides of the solution, then after rearranging, you should be able to solve for $x$.

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    $\begingroup$ With the constraint that $u>0$... $\endgroup$
    – Umberto
    Commented Dec 11, 2014 at 7:11

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