Problem (Algebra):
The equation $2^{x+2} 5^{6-x}=10^{x^{2}}$ has two real solutions. Determine these two solutions.

Solution:
Rewriting the equation, we obtain

$$
\begin{aligned}
2^{x+2} 5^{6-x} & =2^{x^{2}} 5^{x^{2}} \\
1 & =2^{x^{2}} 2^{-2-x} 5^{x^{2}} 5^{x-6} \\
1 & =2^{x^{2}-x-2} 5^{x^{2}+x-6} \\
0 & =\left(x^{2}-x-2\right) \log _{10} 2+\left(x^{2}+x-6\right) \log _{10} 5 \\
0 & =(x-2)(x+1) \log _{10} 2+(x-2)(x+3) \log _{10} 5 \\
0 & =(x-2)\left[(x+1) \log _{10} 2+(x+3) \log _{10} 5\right] \\
0 & =(x-2)\left[\left(\log _{10} 2+\log _{10} 5\right) x+\left(\log _{10} 2+3 \log 105\right)\right] \\
0 & =(x-2)\left[\left(\log _{10} 10\right) x+\log _{10}\left(2 \cdot 5^{3}\right)\right] \\
0 & =(x-2)\left(x+\log _{10} 250\right)
\end{aligned}
$$

Therefore, $x=2$ or $x=-\log _{10} 250$.

We take base 10 logarithms of both sides:

$$
\begin{aligned}
\log _{10}\left(2^{x+2} 5^{6-x}\right) & =\log _{10}\left(10^{x^{2}}\right) \\
\log _{10}\left(2^{x+2}\right)+\log _{10}\left(5^{6-x}\right) & =x^{2} \\
(x+2) \log _{10} 2+(6-x) \log _{10} 5 & =x^{2} \\
x\left(\log _{10} 2-\log _{10} 5\right)+\left(2 \log _{10} 2+6 \log _{10} 5\right) & =x^{2} \\
x^{2}-x\left(\log _{10} 2-\log _{10} 5\right)-\left(2 \log _{10} 2+6 \log _{10} 5\right) & =0
\end{aligned}
$$

Now, $\log _{10} 2+\log _{10} 5=\log _{10} 10=1$ so $\log _{10} 5=1-\log _{10} 2$, so we can simplify the equation to

$$
x^{2}-x\left(2 \log _{10} 2-1\right)-\left(6-4 \log _{10} 2\right)=0
$$

This is a quadratic equation in $x$, so should have at most 2 real solutions.

By the quadratic formula,

$$
\begin{aligned}
x & =\frac{\left(2 \log _{10} 2-1\right) \pm \sqrt{\left(2 \log _{10} 2-1\right)^{2}-4(1)\left(-\left(6-4 \log _{10} 2\right)\right)}}{2(1)} \\
& =\frac{\left(2 \log _{10} 2-1\right) \pm \sqrt{4\left(\log _{10} 2\right)^{2}-4\left(\log _{10} 2\right)+1+24-16 \log _{10} 2}}{2} \\
& =\frac{\left(2 \log _{10} 2-1\right) \pm \sqrt{4\left(\log _{10} 2\right)^{2}-20\left(\log _{10} 2\right)+25}}{2} \\
& =\frac{\left(2 \log _{10} 2-1\right) \pm \sqrt{\left(2 \log _{10} 2-5\right)^{2}}}{2} \\
& =\frac{\left(2 \log _{10} 2-1\right) \pm\left(5-2 \log _{10} 2\right)}{2}
\end{aligned}
$$

since $5-2 \log _{10} 2>0$.

Therefore,

$$
x=\frac{\left(2 \log _{10} 2-1\right)+\left(5-2 \log _{10} 2\right)}{2}=\frac{4}{2}=2
$$

or

$$
x=\frac{\left(2 \log _{10} 2-1\right)-\left(5-2 \log _{10} 2\right)}{2}=\frac{4 \log _{10} 2-6}{2}=2 \log _{10} 2-3
$$

(Note that at any point, we could have used a calculator to convert to decimal approximations and solve.)

Final answer: $2,-\log _{10} 250$