Difference between revisions of "1982 AHSME Problems/Problem 17"
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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> and <math>1/9</math>. Setting these equal to <math>3^x</math>, we can immediately see that there are <math>\boxed{2}</math> real values of <math>x</math> that satisfy the equation. | 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> and <math>1/9</math>. Setting these equal to <math>3^x</math>, we can immediately see that there are <math>\boxed{2}</math> real values of <math>x</math> that satisfy the equation. | ||
Revision as of 11:39, 9 June 2016
Solution
Let 
. Then the preceding equation can be expressed as the quadratic, ![\[9a^2-28a+3 = 0\]](http://latex.artofproblemsolving.com/4/2/2/422ab93927f22c8a9b3a9f52cbfa13901dc24fe1.png)
Solving the quadratic yields the roots 
and 
. Setting these equal to 
, we can immediately see that there are 
real values of 
that satisfy the equation.
