# Difference between revisions of "1982 AHSME Problems/Problem 17"

(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...") |
|||

(One intermediate revision by one other user not shown) | |||

Line 1: | Line 1: | ||

+ | ==Problem== | ||

+ | How many real numbers <math>x</math> satisfy the equation <math>3^{2x+2}-3^{x+3}-3^{x}+3=0</math>? | ||

+ | |||

+ | <math>\text {(A)} 0 \qquad \text {(B)} 1 \qquad \text {(C)} 2 \qquad \text {(D)} 3 \qquad \text {(E)} 4</math> | ||

+ | |||

+ | ==Solution== | ||

+ | |||

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. |

## Latest revision as of 22:02, 16 June 2021

## Problem

How many real numbers satisfy the equation ?

## Solution

Let . Then the preceding equation can be expressed as the quadratic, 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.