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Logarithms: Solving for T


Date: 01/30/98 at 21:01:04
From: Matt Hubble
Subject: I need help solving a math problem

My wife is currently taking a finance course and we are both stuck on
a simple algebra problem that neither one of us has done for years 
since we have completed all of our schooling. The problem works out
as follows:
                T
  350  =  100  x     (1.12)

We have no problem getting to this step:

                T
  3.5  =  (1.12)

Our problem is that neither one of us can remember how to solve for
T at this point. We came across your web site and are hoping you can 
help us out.

Thanks,
Matt


Date: 02/05/98 at 11:37:36
From: Doctor Joe
Subject: Re: I need help solving a math problem

Dear Matt,

To solve your equation, you need to make use of the logarithm 
function.

A small recollection of what logarithm is:

When we encounter expressions like:

(1) 100 = 10^2  (read as 10 to the power of 2)

(2) 5 = 25^(1/2) (read as 25 to the power of half, meaning square  
    roots)

(3) 0.1 = 10^(-1)

we are dealing with indices.

We see that the numbers 10, 25, 10 in each of the above examples are 
the base numbers, often called the base.

The numbers 2, 1/2, -1 are indices, or powers.

How can we express the indices in terms of the numbers on the left of 
the equality and the bases?

In other words, in example 1, how do we write 2 in terms of 100 and 
10; equivalently, how do we express 1/2 in terms of 5 and 25?

We are going to use a notation to resolve this problem of denotation.

Suppose in general that if a is a positive number not equal to 1, and 
there are numbers x and y (positive) such that:

                     a^x = y,

we may define the logarithm of y based a, denoted by

                     log_a (y)

to be equal to x.

In other words, we express the exponent/index x in terms of the number 
y and the base a.

One can show that the log_a is more than a denotation. To be more 
exact, if we write f(x) = log_a (x), then log_a is actually a 
function.

Let's look at two crucial rules that you may find helpful:

Rule 1: Let a be a positive number not equal to 1. Suppose we have two 
        real numbers x and y (y positive). Then,

        log_a (y^x) = x log_a (y).

In words, it simply means the power of y may be brought "down" as a 
multiple of the log of y.

Rule 2: Let a and b be bases (they can only be positive, not equal 
        to 1).

Suppose y is a positive real number.

                          log_a (y)
    Then,    log_b (y) = ----------- .
                          log_a (b)

This is frequently called the change base formula.

The proofs of these two rules may be found in many college books and 
references.  I shall skip the proofs.

Let me show you how these 2 rules may be used to solve problems of 
this sort.

Example 1.

Solve for x in the equation:  2^x = 4.

Solution:  Besides trial and error, we can use the two rules to show 
that the solution is 2.

           2^x = 4

        =>  log_2 (2^x) = log_2 (4)    [applying log_2 on both sides]
        => x log_2 (2)  = log_2 (2^2)
        => x log_2 (2)  = 2 log_2 (2)

At this point, we want to evaluate the expression log_2 (2).

Unwinding the definition, let y = log_2 (2), then,

                     2 ^ y = 2.

This means that y = 1.

{In general, if a is a positive number not equal to 1, log_a (a) = 1.}

Back to the problem, so...

                 x = 2.

Example 2.

Solve for y in the equation:  24 = 12^y.

Solution:  Apply log_10 on both sides,

           log_10 (24) = log_10 (12^y)

The reason why I used log_10 is because 24 and 12 are not rational 
indices of each other (i.e. they are not very nice numbers). However, 
in the example 1, I can also choose 10 as the base. Another reason why 
I choose log_10 is that log_10 (a) may be found by using a Scientific 
Calculator.

Back to the problem:

          log_10 (24) = y log_10 (12)

                log_10 (24)      1.3802
      =>  y = --------------- =  ------ = 1.279 (3 decimal places)
                log_10 (12)      1.0792

* At this point, think about Rule 2.  Note that I have not used it 
anywhere.  In fact, just using Rule 1 suffices.

Now, you may use the same method applied in example 2 to solve for T, 
bearing in mind these few steps:

   Step 1. Applying log_10 on both sides of the equation.

   Step 2. Make the unknown the subject of the formula.

   Step 3. Use the calculator to evaluate the answer.

Check that your answer for T is 11.054 (up to 3 decimal places).

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

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