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This is the equation, but the result is different from wolframalpha:

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

WolframAlpha: $x=0$

Where is the error?

Martin Sleziak
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J.Doe
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    The error is in the step $2^x+2^{2x}=2^1\implies x+2x=1$ which is not true. You can't just take out the exponents this way. – Kenny Lau Jun 25 '16 at 07:40
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    To do this question, let $u=2^x$. Then, $4^x=(2^x)^2=u^2$. Then, the equation becomes $u+u^2=2$, which is a quadratic equation. – Kenny Lau Jun 25 '16 at 07:40
  • why not? Aren't them in the same base? – J.Doe Jun 25 '16 at 07:40
  • If there is only one term, i.e. $2^m=2^n$, then you can say $m=n$. The principle is by taking the log of both sides, i.e. $\log_2(2^m)=\log_2(2^n)$. – Kenny Lau Jun 25 '16 at 07:41
  • See http://math.stackexchange.com/questions/1827123/equation-52x1-621-times10x-100-times4x?lq=1 OR http://math.stackexchange.com/questions/1804249/solve-equation-4x-3-cdot6x2-cdot9x-0?lq=1 – lab bhattacharjee Jun 25 '16 at 07:42
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    However, if you have the form $2^a+2^b=2^c$, then when you try to take the log of both sides, you would get $\log_2(2^a+2^b)=c$, but $\log_2(2^a+2^b)\ne\log_2(2^a)+\log_2(2^b)$ – Kenny Lau Jun 25 '16 at 07:42
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    $2^x\cdot 2^{2x}=2^1$ implies that $x+2x=1$. Exponentiation turns a sum to a product (and contrariwise taking logarithms turns a product into a sum). There is no formula for $\log(a+b)$ in general. – Jyrki Lahtonen Jun 25 '16 at 07:42
  • ok, with substitution I get $u_1=1$ and $u_2=-2$ – J.Doe Jun 25 '16 at 07:42
  • Then $2^{x_1}=1$ and $2^{x_2}=-2$. – Kenny Lau Jun 25 '16 at 07:43

4 Answers4

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$$2^x+2^{2x}=2$$

Now put $2^x=t$

$$t+t^{2}=2$$

$$t^{2}+t-2=0$$

$$(t-1)(t+2)=0$$

Thus $t=1$ or $t=-2$

$2^x=1$ or $2^x=-2$

Since $2^x>0 $ for all real $x$ , $2^x=1=2^0$

Therefore $x=0$

Angelo Mark
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$a^x \cdot a^y=a^{x+y}$ is the identity
you are using $a^x+a^y=a^{x+y}$ which is not correct

Martin Sleziak
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Vaibhav Jain
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2

$2^x + 4^x= 2$ $\Rightarrow$ $2^x (1 + 2^x ) = 2$ $\Rightarrow $ $1 + 2^x = 2 ^{1 - x}$ $\Rightarrow$ $1 + 2^x = 2 ^{- x} \times 2$

Now Set $ y = 2^x$; then we have

$1 + y = y^{-1} \times 2 $ $\Rightarrow$ $y^2 + y -2 = 0$. Which has solutions $y = 1$ and $y = -2$.

$y = -2 $ is unacceptable, because $y = 2^x$ is a positive function. So $y = 2^x = 0$ is acceptable and will give us $ x = 0$. And we are done!

Farrokh
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we have $$2^{2x}+2^x-2=0$$ and with $$2^x=t$$ you will get $$t^2+t-2=0$$ a quadratic equation to solve.

Dr. Sonnhard Graubner
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