2006 AMC 10A Problem 6

The exponents $14$ and $7$ are related ($14 = 2 \times 7$), so try to make both sides a 7th power.

1. Squaring first and then taking the 7th power is the same as taking the 14th power. So
$$(7x)^{14} = \big((7x)^2\big)^7 = (49x^2)^7.$$
2. The equation becomes
$$(49x^2)^7 = (14x)^7.$$
3. An odd power like the 7th never gives the same answer for two different numbers (for example $2^7 = 128$ and $(-2)^7 = -128$). So the two bases must be equal:
$$49x^2 = 14x.$$
4. The problem says $x$ is not zero, so we may divide both sides by $7x$:
$$7x = 2, \qquad x = \frac{2}{7}.$$
5. Check: with $x = \frac27$, we get $7x = 2$ and $14x = 4$. Is $2^{14} = 4^7$? Yes, because $4^7 = (2^2)^7 = 2^{14}$.

The answer is $\boxed{\textbf{(B)}\ \frac{2}{7}}$.

Equations with powers get easy when both sides have the same exponent (or the same base). Here $14$ is twice $7$, so one side could be rewritten as a 7th power, and the problem turned into a simple equation. The words "non-zero" are a hint that $x = 0$ would also work and has to be thrown out.

Test the choices. For $x = \frac27$ the two bases are $7x = 2$ and $14x = 4$, and $2^{14} = 4^7$ is true. For $x = 1$ you would need $7^{14} = 14^7$, that is $49^7 = 14^7$, which is false; the other choices fail the same way.