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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?

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    $\begingroup$ 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. $\endgroup$
    – Kenny Lau
    Jun 25, 2016 at 7:40
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    $\begingroup$ 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. $\endgroup$
    – Kenny Lau
    Jun 25, 2016 at 7:40
  • $\begingroup$ why not? Aren't them in the same base? $\endgroup$
    – J.Doe
    Jun 25, 2016 at 7:40
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    $\begingroup$ 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 $\log_2(2^a+2^b)\ne\log_2(2^a)+\log_2(2^b)$ $\endgroup$
    – Kenny Lau
    Jun 25, 2016 at 7:42
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    $\begingroup$ $2^x\cdot 2^{2x}=2^1$ 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. $\endgroup$ Jun 25, 2016 at 7:42

4 Answers 4

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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 $x=0$

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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

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$2^x + 4^x= 2$ $\Rightarrow$ $2^x (1 + 2^x ) = 2$ $\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!

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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.

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