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Solving Logistic Equations for a Variable in the Exponent

Date: 05/03/98 at 12:24:06
From: Paula-Maree Cavenett
Subject: (no subject)

These questions pertain to logistic growth.

1.) The formula for Q is as follows:

    Q = U/[(1 + exp(-a - (bt))]

We have:
    a = -4.38
    b =  0.06
    U =  663100
    Q =  151900
    t =  ?

I need to know how to get t.

2.) Given:
    a = -4.36
    b =  0.86
    U =  4033333
    q =  200000
    t =  ?

Our formula for q is as follows:

    q = [(UB) exp(-a - (bt))]/[1 + exp(-a - (bt))^2]

I need to solve for t again.

Can you please help?

Date: 05/03/98 at 13:07:28
From: Doctor Sam
Subject: Re: (no subject)

Hi Paula-Maree!

I'll try to show you the steps needed to solve for t in equations like 
the ones that you provided.

Problem 1:

   Q = U/[(1 + exp(-a - (bt))]

First, multiply both sides of the equation by the denominator 
1 + exp(-a - bt) in order to clear fractions:

   Q(1 + exp(-a - bt)) = U

Our goal is to isolate the t. This will take several steps. Divide 
both sides by Q:

    1 + exp(-a - bt) = U/Q

Next, subtract 1 from both sides:

    exp(-a - bt) = U/Q  - 1

Now we need to get at the t, which is actually part of an exponent of 
e. The tool for "unwrapping" the exp function is the natural logarithm 
function ln(x). In general, ln[exp(x)] = x, and exp[ln(x)] = x, too.

So now we "take natural logarithms" of both sides of the equation:

   ln[exp(-a - bt)]  = ln[U/Q - 1]

which simplifies to

   -a - bt = ln[U/Q  - 1]

Now just add "a" to both sides and then divide through by "-b":

   -bt = ln[U/Q  - 1] + a

        ln[U/Q  - 1] + a
   t =  ----------------

Problem 2:

Solving for t in q = [(UB) exp(-a - (bt))]/[1 + exp(-a - (bt))^2] will 
be more complicated, because t occurs in both the numerator and the 
denominator of the right hand side. We begin the same way, by clearing 

   [1 + exp(-a - (bt))^2] q = [(UB) exp(-a - (bt))]

In this equation, the exp function appears twice, and once it is being 
squared. This makes me think about quadratic equations. To help myself 
think about it, I am temporarily going to write x = exp(-a - bt) to 
change this equation into:

   (1 + x^2) q = (UB) x  

which looks a lot easier.

This is a quadratic equation which we can solve by multiplying it out, 
gathering all the terms on the same side of the equation, and applying 
the quadratic formula:

   q +  qx^2 = UBx
   qx^2 - UBx + q = 0

The quadratic formula gives two solutions:

   x = [UB + sqrt((UB)^2 - 4q^2)]/2q   and
   x = [UB - sqrt((UB)^2 - 4q^2)]/2q

Now is the time to replace x by exp(-a - bt). We have two equations to 
solve instead of just one, but the two solutions are very similar. 

   exp(-a - bt) = [UB + sqrt((UB)^2 - 4q^2)]/2q 

Since we have exp(of something) = "something else," we can take 
natural logarithms of both sides of the equation, as we did in the 
first problem:

   ln[exp(-a - bt)] = ln{[UB + sqrt((UB)^2 - 4q^2)]/2q}

           -a - bt  = ln{[UB + sqrt((UB)^2 - 4q^2)]/2q}

And once again add "a" and divide through by "-b":

               -bt  = a + ln{[UB + sqrt((UB)^2 - 4q^2)]/2q}

                      ln{[UB + sqrt((UB)^2 - 4q^2)]/2q})
                  t = -----------------------------------

There is another solution, where the square root is subtracted.  

One caution about all the solutions: the domain of ln(x) includes only 
positive numbers. It is meaningless to take ln(x) if x < 0. So in both 
problems, the solution will only work for values of the many variables 
that produce a positive number inside the natural logarithm function.

I hope that helps.

-Doctor Sam, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
Associated Topics:
High School Basic Algebra
High School Logs

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