2013 AMC 10A Problem 8

The powers are huge, so the actual numbers cannot matter. Only how they compare matters.

1. Try small numbers first. Replace $2014$ and $2012$ by $4$ and $2$:
$$\frac{2^4 + 2^2}{2^4 - 2^2} = \frac{16 + 4}{16 - 4} = \frac{20}{12} = \frac53.$$
Notice $16$ is $4$ times $4$. The top is $4 \times (4 + 1)$ and the bottom is $4 \times (4 - 1)$.
2. Do the same with the real numbers. Multiplying $2^{2012}$ by $2^2 = 4$ adds $2$ to the exponent, so
$$2^{2014} = 4 \times 2^{2012}.$$
3. Pull out the common part. Both the top and the bottom contain $2^{2012}$:
$$\frac{4 \times 2^{2012} + 2^{2012}}{4 \times 2^{2012} - 2^{2012}} = \frac{2^{2012} \times (4 + 1)}{2^{2012} \times (4 - 1)}.$$
4. Cancel. Now $2^{2012}$ is multiplied on top and bottom, so it cancels, leaving $\dfrac{5}{3}$.

The answer is $\boxed{\textbf{(C)}\ \frac{5}{3}}$.

When you add or subtract powers of the same number, pull out the smallest power first. What remains is small and easy. You may only cancel something from top and bottom when it multiplies everything, which is why Step 3 comes before Step 4. Trying a small case (Step 1) is a good way to see the pattern and to check the answer.

Write $2^{2014} = 2^{2012} \cdot 2^2 = 4 \cdot 2^{2012}$. Then both the numerator and the denominator have the common factor $2^{2012}$:
$$\frac{2^{2014} + 2^{2012}}{2^{2014} - 2^{2012}} = \frac{2^{2012}(4 + 1)}{2^{2012}(4 - 1)} = \frac{5}{3}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{5}{3}}$.