TITLE: Exponential Simultaneous Equations
QUESTION [8 upvotes]: 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.

REPLY [9 votes]: 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
$$x=1\text{ or }x=3.$$

REPLY [6 votes]: Set $z=2^x$ to obtain $z^2-10z+16=0$ and solve the quadratic

REPLY [3 votes]: 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
$$2^x,2^y=2,8.$$

REPLY [2 votes]: 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:
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=1,\ 3}}$$