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A function $y(x)$ is defined as $$ 2^y+2^x=2 $$ The question is about finding it's domain. Pretty simple. By observing the function I could say all the negative numbers are in the domain. But, I think $0$ is included in the domain because the function is defined at $0$ . The text book says $0$ is not included. How is that?

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  • $\begingroup$ I think $0$ can be included $\endgroup$ – Empty May 5 '15 at 3:03
  • $\begingroup$ I think $x=0$ is okay as well since there is a solution ($y(0) = 1$). The only issue is whether such a $y$ is unique. $\endgroup$ – Cameron Williams May 5 '15 at 3:10
  • $\begingroup$ Isn't the value of y zero when x equals zero? $\endgroup$ – GrandAlpha May 5 '15 at 3:13
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Certainly $0$ is in the domain; when $x=0$, we have $2^y + 2^0=2$, for which there is a unique solution.

I suspect that there is a typo in the book and they intended to say that $x=1$ is not included. When $x=1$, then $2^y+2^1=2$, but that means $2^y=0$, which is impossible. The domain is $(-\infty,1)$.

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$\bf{My\; Solution}$ Given $$\displaystyle 2^x+2^y = 1\Rightarrow 2^y=(1-2^x)$$

Now for $x=0\;,$ We get $2^0+2^y=2\Rightarrow 2^y=1\Rightarrow y=0$

Now taking $\log_{2}$ on both side, We get $$\displaystyle \log_{2}(2)^y = \log_{2}(1-2^x)$$

So we get $$y=\log_{2}(1-2^x)\;,$$ Which is defined when $$(1-2^x)>0$$

So $$2^x-1<0\Rightarrow 2^x<2^0\Rightarrow x<0$$

So we get Domain $$x\in \left(-\infty\;,0\right]$$

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  • $\begingroup$ given 2^x+2^y=2 $\endgroup$ – GrandAlpha May 5 '15 at 3:37

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