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Solving Logarithms with the Quotient Rule


Date: 11/11/98 at 20:22:25
From: Tierney Sparks
Subject: Logarithms and dividing them

Dr. Math, 

I have a test in a few days and really need your help. This problem is 
bound to be on the test:

   log(base5) (9x+2) - log(base5) (3x+8) = 2

My first step was to use the quotient property, but I got lost. Can 
you please help?


Date: 11/12/98 at 08:25:23
From: Doctor Rick
Subject: Re: Logarithms and dividing them

Hi, Tierney. 

Before I answer your question, I must clear up some terminology. I 
think this is your problem:

   log (9x + 2) - log (3x + 8) = 2
      5              5

When you say this out loud, it is "log (base 5) OF (9x + 2) etc.", not 
"log (base 5) MULTIPLIED BY (9x + 2) etc." The log is not something 
that you can multiply by a number. It is an operation or function, 
something that you do to a number. It's like the square root in this 
respect: you don't do "the square root times 2" but "the square root 
of 2".

Next, you are correct that you should apply the "quotient property." 
Remember, the quotient property is this:

   log(a) - log(b) = log(a/b)

You need to put the quotient inside the log function, like this:

         9x + 2
   log ( ------ ) = 2
      5  3x + 8

I will tell you the second step, and then I think you will know what to 
do. You can make each side of the equation the exponent of 5, and use 
this property of exponents:

   log (x) = y means
      n
 
   n^y     = x

So we do this first:

         9x + 2
   log ( ------ ) = 2  means
      5  3x + 8      

   5^2 = (9x + 2)/(3x + 8) 

and it simplifies to this:

   9x + 2
   ------ = 25
   3x + 8

Now it's all yours!

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

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