Difference between revisions of "1965 AHSME Problems/Problem 1"
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Take the logarithm with a base of <math>2</math> to both sides, resulting in the equation <math>2x^2-7x+5 = 0</math>. Factoring results in <math>(2x-5)(x-1) = 0</math>, so there are <math>\boxed{\textbf{(C) } 2}</math> real solutions. | Take the logarithm with a base of <math>2</math> to both sides, resulting in the equation <math>2x^2-7x+5 = 0</math>. Factoring results in <math>(2x-5)(x-1) = 0</math>, so there are <math>\boxed{\textbf{(C) } 2}</math> real solutions. | ||
− | == | + | ==Solution 2== |
Notice that <math>a^0=1, a>0</math>. So <math>2^0=1</math>. So <math>2x^2-7x+5=0</math>. Evaluating the discriminant, we see that it is equal to <math>7^2-4*2*5=9</math>. So this means that the equation has two real solutions. Therefore, select <math>\fbox{C}</math>. | Notice that <math>a^0=1, a>0</math>. So <math>2^0=1</math>. So <math>2x^2-7x+5=0</math>. Evaluating the discriminant, we see that it is equal to <math>7^2-4*2*5=9</math>. So this means that the equation has two real solutions. Therefore, select <math>\fbox{C}</math>. | ||
Revision as of 09:09, 18 July 2024
Contents
Problem
The number of real values of satisfying the equation
is:
Solution 1
Solution by e_power_pi_times_i
Take the logarithm with a base of to both sides, resulting in the equation
. Factoring results in
, so there are
real solutions.
Solution 2
Notice that . So
. So
. Evaluating the discriminant, we see that it is equal to
. So this means that the equation has two real solutions. Therefore, select
.
~hastapasta
See Also
1965 AHSME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.