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### Solving Logarithms with the Quotient Rule

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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
```

```
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

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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