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1982 AHSME Problems/Problem 17

Revision as of 10:39, 9 June 2016 by Dot22 (talk | contribs) (Created page with "Let <math>a = 3^x</math>. Then the preceding equation can be expressed as the quadratic, <cmath>9a^2-28a+3 = 0</cmath> Solving the quadratic yields the roots <math>3</math> an...")
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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.

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