1965 AHSME Problems/Problem 1
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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.
Solution2
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 | |
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All AHSME Problems and Solutions |
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