Solving Logarithm EquationsDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/