# 1982 AHSME Problems/Problem 17

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Let $a = 3^x$. Then the preceding equation can be expressed as the quadratic, $\[9a^2-28a+3 = 0\]$ Solving the quadratic yields the roots $3$ and $1/9$. Setting these equal to $3^x$, we can immediately see that there are $\boxed{2}$ real values of $x$ that satisfy the equation.