2006 AMC 10A Problem 6

Rewrite the left side as a seventh power: $(7x)^{14} = \left((7x)^2\right)^7 = (49x^2)^7$. The equation is now
$$(49x^2)^7 = (14x)^7.$$
Seventh powers of real numbers agree only when the bases agree (an odd power never maps two different reals to the same value), so
$$49x^2 = 14x \quad\Longrightarrow\quad 7x(7x - 2) = 0.$$
The root $x = 0$ is excluded, leaving $7x = 2$, i.e. $x = \dfrac{2}{7}$.

Check: $7x = 2$ and $14x = 4$, and $2^{14} = 4^7$ indeed.

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

Exponent equations become manageable when both sides are expressed with the same exponent (or the same base). Here the exponents $14$ and $7$ share the factor $7$, so peeling off a common seventh power reduces the problem to a quadratic. The "non-zero" condition warns that $x = 0$ will show up and must be discarded.

Test the choices: $x = \tfrac{2}{7}$ makes the bases $2$ and $4$, and $2^{14} = (2^2)^7 = 4^7$. The other choices give bases like $1$ and $2$ or $7$ and $14$, which clearly fail.

• Cancelling the $7$ and $14$ as if they were bases, e.g. concluding $7x = 14x$ or $7x = 2x$, or dividing by $x$ carelessly and getting $x = 7$ (choice (D)).
• Taking $(7x)^{14} = 7x^{14}$, forgetting that the coefficient is raised to the power too.