1972 AHSME Problems/Problem 6

Problem 6

If $3^{2x}+9=10\left(3^{x}\right)$, then the value of $(x^2+1)$ is

$\textbf{(A) }1\text{ only}\qquad \textbf{(B) }5\text{ only}\qquad \textbf{(C) }1\text{ or }5\qquad \textbf{(D) }2\qquad \textbf{(E) }10$

Solution

We can rearrange to get $3^{2x}-10\left(3^{x}\right)+9=0$. Factor or use the quadratic formula to find that $3^{x}= 9\text{ or }1$, which makes $x = 2\text{ or }0$ and our answer \[\boxed{\textbf{(C) }1\text{ or }5}.\]