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### 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/
```
Associated Topics:
High School Logs

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