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Challenge 5: Since x, y and z are all the bases of logarithms, they must all be positive but not equal to 1. Since z > 0, either x , y > 1 or 0 < x , y < 1 in order for to be positive. Similarly, the pair x and z and the pair y and z must also be both greater than 1 or both in between 0 and 1. Combining these, x , y and z must be either all greater than 1 or all in between 0 and 1. . Thus, so that xyz = 1. This is a contradiction since we determined above that either all three of x , y , and z are greater than 1, in which case xyz would be greater than 1, or all three of x , y , and z are in between 0 and 1, in which case xyz would be less than 1. Therefore, there is no solution. However, using the change of base formula, we have . From the first of these, but combining the last two gives , so that we also have

Question 6:

Challenge 6:

is quadratic in the variable y = cos x , and if x is in

, then y =

The equation

cos x is in the interval [0,1]. Thus, we may concentrate on solving

with two unique solu

tions in [0,1] when a > 0.

First, for the solutions to be real and unique, the Quadratic Formula says we must have

Next, in order for the smaller solution to be in [0,1], we must have

, which re

. Since the right - hand side is positive, b must be negative. Furthermore, squaring

quires

. Since a > 0 we must have c ≥ 0.

both sides gives

or

For the larger solution, we must have

.. Thus,

,

or

, which gives us two constraints. First, since the left - hand side is positive, we must

have b + 2 a > 0 so that, b > - 2 a . We may also square both sides to get

b 2 - 4 ac ≤ b 2 + 4 ab + 4a 2 - 4 ac ≤ 4 ab + 4 a 2

c ≥ - b - a.

Therefore, combining these constraints, we have, given a > 0, we must have - 2 a < b < 0 and b 2 – 4 ac > 0, c ≥ 0, and c ≥ - b - a , which may be combined to get, min { - b – a , 0} ≤ c ≤

Virginia Mathematics Teacher vol. 48, no. 1

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