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Solving Logarithm Equations


Date: 04/29/2001 at 00:15:08
From: David Thomas
Subject: Logarithms

Solve for y:

     log_7(x) / log_7(4) - log_7(y) = 1

where log_7 means the logarithm taken base 7.

I've asked several high school math teachers for help. We've looked in 
several math textbooks. The texts seem to only give work on the three 
basic laws of logarithms and not deal with the quotient in this one. 
There is an exercise in my textbook with an answer, but no explanation 
on the answer and no examples like this question. Here is the exercise 
and the solution:

     2log_5(x) / log_5(3) - log_5(y) = 2

Solution:  y = (1/25)x^(2/log _5(3))


Date: 04/29/2001 at 10:20:36
From: Doctor Scott
Subject: Re: Logarithms

Hi David!

Let's run through the steps for the sample problem that you provided 
and hope that will give you enough to figure out your exercise.

Since all of the logarithms are base 5, I am going to leave that part 
out until the last step (when we use that fact) so that it looks 
"neater" when I write it.

Original problem:

     2log(x)
     ------- - log(y) = 2         log means log base 5
      log(3)

     2log(x)
     ------- - 2 = log(y)         move the "y" term to the left
      log(3)

          2log(x)
        ( ------- - 2 )
           log(3)
     5 ^                  = y     use exponents to solve for y

Now we have at least isolated y. We need now to simplify the 
expression on the left. To do so we will use the properties of 
exponents:
                              

      (m+n)    m  n       m/n     m   1/n
     a      = a  a   and a    = (a  )

So,

             2log(x)                2log(x)
           ( ------- - 2 )        ( ------- )
              log(3)                 log(3)        -2
        5 ^                 =  5 ^            * 5 ^

             2log(x)
           ( ------- )
              log(3)        -2
     =  5 ^            * 5 ^

            2log(x)   1/log(3)
     =  (5 ^        )          * (1/25)
          
            log(x^2)   1/log(3)
     =  (5 ^         )          * (1/25)

Now using logarithms, (remember that log means log base 5), 

         log(x^2)
     5 ^           =  x^2

so now we have:

            1/log(3)  
     (x^2)^          * (1/25)

and that's just:

                2
              ------
              log(3)  
     (1/25) x

Whew!  Hope that helps!

- Doctor Scott, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Logs

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