1982 AHSME Problems/Problem 17

Solution

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.

Page

Toolbox

Search

1982 AHSME Problems/Problem 17

Solution