Equations with Logarithms
Date: 04/18/97 at 14:02:12 From: Daniel Osman Subject: logarithms Here is my problem: x^3 - (3/2)x^2 + 5/2 = log (base 1/4) (m) Find those values of 'm' for which this equation has 3 different solutions. I only know that this equation can be written like this: (1/4)^(x^3 - (3/2)x^2 + 5/2) = m
Date: 04/19/97 at 08:12:05 From: Doctor Anthony Subject: Re: logarithms We first find the solutions of x^3 - (3/2)x^2 + 5/2 = 0 If x = -1, -1 - 3/2 + 5/2 = 0, so x+1 is a factor. Dividing out by x+1, we get x^2 - (5/2)x + 5/2, which has no more real factors. So m = 1 would not give 3 different solutions. What we require are values of m such that: x^3 - (3/2)x^2 + 5/2 = log(base 1/4) (m) has three real roots. We write the equation as f(x) = x^3 - (3/2)x^2 + 5/2 - k where k is the value of log(base 1/4) (m). Now there will be three real roots to f(x) = 0 if the turning points of this cubic lie on either side of the x axis. So we adjust the value of k to ensure that this happens. f'(x) = 3x^2 - 2(3/2)x = 3x^2 - 3x = 3x(x-1) = 0 for turning points. So turning points occur when x = 0 and x = 1. x = 0 will be the maximum turning point, so f(x) must be positive at x = 0. This means 5/2 - k > 0 or k < 5/2 x = 1 is the minimum turning point, and this must lie below the x axis, i.e. with f(x) < 0. f(1) = 1 - 3/2 + 5/2 - k < 0 2 - k < 0 k > 2 So for the cubic to have 3 real roots, we require 2 < k < 5/2. This means 2 < log(base 1/4) m < 5/2 (1/4)^2 > m > (1/4)^(2.5) note we reverse direction of inequality .0625 > m > 0.03125 Any value of m in this range will give three real roots to the cubic, and so three real roots to the original equation. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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