1965 AHSME Problems/Problem 1
Contents
[hide]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) | ||
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