Solving Logarithms with the Quotient RuleDate: 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/ |
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