# 1978 AHSME Problems/Problem 17

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## Problem 17

If $k$ is a positive number and $f$ is a function such that, for every positive number $x$, $\left[f(x^2+1)\right]^{\sqrt{x}}=k$; then, for every positive number $y$, $\left[f(\frac{9+y^2}{y^2})\right]^{\sqrt{\frac{12}{y}}}$ is equal to $\textbf{(A) }\sqrt{k}\qquad \textbf{(B) }2k\qquad \textbf{(C) }k\sqrt{k}\qquad \textbf{(D) }k^2\qquad \textbf{(E) }y\sqrt{k}$

We are given that $$[f(x^2 + 1)]^{\sqrt(x)} = k$$ We can rewrite $\frac{9+y^2}{y^2}$ as $\frac{9}{y^2} + 1$ Thus, our function is now $$[f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{12}{y}}} = k$$ $$\Rrightarrow[f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y} \cdot 4}} = k$$ $$\Rrightarrow([f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y}}})^{\sqrt{4}} = (k)^{\sqrt{4}}$$ $$\Rrightarrow([f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y}}})^{2} = (k)^{2} = k^2$$ $$\boxed{D}$$

~JustinLee2017

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