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Inverse Hyperbolic Cosine in Terms of ln

Date: 05/27/2003 at 07:37:17
From: Anonymous 
Subject: Logarithm equations without a rule to govern them!

What is t in the equation ln(e^0.1t + 9e^-0.1t) = ln10? 

There are no rules governing an "expansion" of the ln, or even how to 
do ln9e^-0.1t

I know from just subsituitng some numbers that ln(e^0.1t + 9e^-0.1t) 
does not = lne^0.1t + lne^-0.1t.

Date: 05/27/2003 at 08:26:08
From: Doctor Mitteldorf
Subject: Re: Logarithm equations without a rule to govern them!

Hi there,

I understand your frustration, but I suggest that it derives from a
simple truth that none of your instructors has yet revealed to you: 
The vast majority of equations that you might write down do not admit 
of an algebraic solution.

This fact is occluded in the educational process by the 
(understandable) tendency of teachers to focus on those problems that
DO have algebraic solutions. This leaves most students with the
impression that ALL algebraic equations have algebraic solutions, and
consequently that a failure to discover a solution for a given 
equation reflects poorly on their own ability as mathematicians.

In fact, the equation that you have written

> ln(e^0.1t + 9e^-0.1t) = ln10

cannot be solved for t.  You can raise e to the power of each side

   e^0.1t + 9e^-0.1t = 10

But this is as far as you can go. One way is to guess values of t, 
evaluate the LHS, and see how close you get to 10, then use the
disparity to help with your next guess. This is called "Newton's
Method" or the Newton-Raphson method.

   Inventing an Operation to Solve x^x = y 

If you didn't have the 9, however, there would be a combination of
symbols that is conventionally notated as the "hyperbolic cosine" or
cosh, defined as 

   cosh(x) = (e^x + e^-x)/2

If you take your equation (without the 9) and raise e to the power of 
each side, you have

   cosh(0.1t) = 20

This gives a solution of the equation as an inverse hyperbolic cosine.  
In one sense, this is an answer; but in another sense it just assigns 
a name to an iterative calculation that arises because there is no 
algebraic solution.

The other equation you offer can be solved algebraically:

   ln9e^-0.1t = ln10

  9 * e^-0.1t = 10

      e^-0.1t = 10/9

       e^0.1t = 9/10
        0.1*t = ln(9/10)

            t = 10 * ln(9/10)
- Doctor Mitteldorf, The Math Forum 

Date: 05/27/2003 at 09:50:38
From: Doctor Peterson
Subject: Re: Logarithm equations without a rule to govern them!


I'd like to add a bit to Dr. Mitteldorf's excellent answer.

He transformed your equation to the form

  e^0.1t + 9e^-0.1t = 10

and mentioned that a special function has been defined,

  cosh(x) = (e^x + e^-x)/2

I notice that your equation (with the 9) can be rewritten using the 
hyperbolic cosine:

  1/3 e^0.1t + 3 e^-0.1t = 10/3                  [dividing by 3)

  e^(0.1t - ln(3)) + e^(-0.1t + ln(3)) = 10/3    [moving coefficients]

  [e^(0.1t - ln(3)) + e^(-0.1t + ln(3))]/2 = 5/3 [dividing by 2]

  cosh(0.1t - ln(3)) = 5/3                       [using the function]

  0.1t - ln(3) = cosh^-1(5/3)                    [taking the inverse]

  t = 10[ln(3) + cosh^-1(5/3)]                   [solving]

As he pointed out, cosh is just a new name given to something that 
otherwise would have to be solved by numerical methods, so in a sense 
this is not really an algebraic solution; but the fact that we can use 
algebra to manipulate a new equation into the form of one that has 
been given a name means that we have actually accomplished something.

This is not uncommon in more advanced applications of algebra or 
calculus: we find a whole class of equations that can be solved using 
a single new function (or group of related functions, such as the 
hyperbolic functions), so that by making tables (in the old days) or 
writing a program to evaluate the newly introduced functions, we can 
solve many problems. So a lot of the algebra you do in solving 
problems like this is to transform your problem into the "standard" 
form for its class (once you have recognized what that is). Other 
examples of such special functions used in solving classes of problems 
include the Bessel functions, the Elliptic integrals, and so on. The 
logarithms and the trigonometric functions are more familiar examples.

I'm not sure I would have noticed that cosh could be used to solve 
this problem without Dr. Mitteldorf pointing it out. (I hope I would, 
but...) There can be a lot of "art" involved in this sort of thing; 
it's like being lost at sea and having to recognize by the smell of 
the air which group of tropical islands is nearest, and finding a way 
to head in that direction, rather than just following a compass home. 
When you finally land on Hyperbolia the natives can care for you. It 
would be nice if there were a simple procedure that would always get 
you "home" without having to use what amounts to intuition to see what 
direction to go in; but despite whatever we say to kids to encourage 
them, it remains true that "math is hard"! It's only the basic 
problems (which tend to be taught in school) that are "easy."

Fortunately, in modern life there are plenty of tools to solve 
equations for you numerically, so if you understand the basics, these 
hard parts can be done automatically. We don't all have to be South 
Seas navigators.

- Doctor Peterson, The Math Forum 

Date: 05/27/2003 at 12:05:35
From: Doctor Peterson
Subject: Re: Logarithm equations without a rule to govern them!


Thinking again about this problem, I realized that the inverse 
hyperbolic cosine CAN be expressed algebraically in terms of ln, so 
although our general advice about non-algebraic solutions is valid, 
it is not necessary here. We CAN solve your problem without resorting 
to a calculator that knows cosh^-1.

Here is how to derive an equation for cosh^-1:

Suppose we know that

  cosh(u) = v

That is,

  e^u + e^-u = 2v

Then we can multiply through by e^u:

  e^(2u) + 1 = 2v e^u

This can be written as a quadratic equation in e^u:

  (e^u)^2 - 2v (e^u) + 1 = 0

Now we can solve for e^u using the quadratic formula:

  e^u = [2v +/- sqrt(4v^2 - 4)]/2

      = v +/- sqrt(v^2 - 1)

  u = ln(v +/- sqrt(v^2 - 1)

You can either apply this expression for the inverse hyperbolic 
cosine to what I wrote, or else apply the same technique to solve 
your problem directly, by multiplying your equation (in exponential 
form) by e^(0.1t) and writing it as a quadratic.

I'm sorry it took so long for us to recognize this solution!

Note that, as we said, solutions to problems like this tend to depend 
on knowing the right tricks; you can see that it is easy even for us 
to miss a solution if we get our minds on the wrong track.

- Doctor Peterson, The Math Forum 

Date: 05/28/2003 at 07:11:56
From: Anonymous
Subject: Thank you (Logarithm equations without a rule to govern 

To Drs. Mitteldorf and Peterson, 

You have brightened my day. This site has to be the best thing for 
people like me who are not "south sea navigators" (as Dr. Peterson 
said) but who have to do a bit of mathematics in their respective 
applications (I do medicine - and I've done some physics and more 
basic maths in the past, but I've never done the "higher level stuff," 
only dabbled in from time to time. I'm learning it now as I go. 

I hope everyone thanks you for all your hard work and polite answers! 
A student of the University of 
Queensland, Brisbane, Australia
Associated Topics:
College Trigonometry
High School About Math
High School Trigonometry

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