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?

  • $\begingroup$ 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

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 $(-\infty,1)$.


$\bf{My\; Solution}$ 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 $$x\in \left(-\infty\;,0\right]$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.