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Solve the following simultaneous equations:

$$2^x + 2^y = 10$$

$$x + y = 4$$

Looking at it, it is obvious that the answers are $(3,1)$ and $(1,3)$, however, I was wondering if they could be solved algebraically. Here's my approach:

$$2^x + 2^{4-x} = 10$$

$$2^x + \frac{(2^4)}{(2^x)} = 10$$

$$2^x + \frac {16}{2^x} = 10$$

And this is where I get stuck. Any help will be greatly appreciated, thanks in advance.

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  • $\begingroup$ This is from a high school textbook, in an earlier chapter to the introduction of complex numbers, which implies that both x and y are real numbers. The answer I have stated as obvious is correct, but am struggling to find a solution algebraically. $\endgroup$ Aug 10, 2015 at 9:18
  • $\begingroup$ The short answer: from the form of the equation you created, it's clear that solving for $2^x$ is simpler than solving for $x$. And once you know $2^x$, you can then solve for $x$. $\endgroup$
    – user14972
    Aug 10, 2015 at 15:21

4 Answers 4

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The next step is

$$\left(2^x\right)^2-10\cdot2^x+16=0$$ wich is a quadratic equation in $2^x$.

Then $2^x=5\pm3$ and $$x=1\text{ or }x=3.$$

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Set $z=2^x$ to obtain $z^2-10z+16=0$ and solve the quadratic

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Alternatively,

$$2^x+2^y=10\\2^x\cdot2^y=16.$$

You know the sum and the product, so by $\left(2^x-2^y\right)^2=\left(2^x+2^y\right)^2-4\cdot 2^x\cdot 2^y$, you get

$$2^x,2^y=2,8.$$

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Notice that:

\begin{align*} 2^x+\frac{16}{2^x}={}&10 \\ (2^x)^2+16={}&10\cdot 2^x \\ (2^x)^2-10\cdot 2^x+16={}&0 \\ (2^x)^2-8\cdot 2^x-2\cdot 2^x+16={}&0 \end{align*}

By factorizing we get

\begin{align*} &\phantom{2^x-8}(2^x-8)(2^x-2)=0 \iff{} \\ &{}\iff 2^x-8=0\vee2^x-2=0 \iff{} \\ &{}\iff 2^x=8\vee2^x=2\iff {} \\ &{}\iff2^x=2^3\vee2^x=2^1\iff{} \\ &{}\iff x=3\vee x=1. \end{align*} Hence, we get that the solutions are: $$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=1,\ 3}}$$

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  • $\begingroup$ Could someone please explain the reason for what is wrong in the answer if you don't mind. $\endgroup$ Aug 10, 2015 at 12:03
  • $\begingroup$ The & line is not really right. It should be an or. I'd put it this way: $(2^x-8)(2^x-2)=0\iff 2^x=8 \vee 2^x=2 \iff x=3 \vee x=1$. $\endgroup$
    – MickG
    Aug 10, 2015 at 13:32
  • $\begingroup$ OK, thanks for your comment. $\endgroup$ Aug 10, 2015 at 13:34
  • $\begingroup$ To really dot the i's and cross the t's, one would have to say "since $2^x\neq0$ I can multiply" and that the first $\iff$ in my comment above is by the zero product property of the reals. That is all I could object to your answer. $\endgroup$
    – MickG
    Aug 10, 2015 at 13:35
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    $\begingroup$ Also, I suggest you look into multi-lined displays in LaTeX, because using multiple subsequent $$ is generally not a great idea. See this and this for example. $\endgroup$
    – MickG
    Aug 10, 2015 at 13:52

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