Assignment 61
Ln(x) is simply Log_{b}x. Ln simply means that the base is e. If you use base e, you are working with natural logarithms. If you use base 10, you are working with common logarithms. Your calculator has only two log functions. One with base 10 (LOG) and one with base e (LN). However, you can find logarithms for any legal base by using what I like to call the Crazy Base Theorem, a description offered by Allan Gunther, former great math teacher at Cate. (Most books call this theorem the Change-of-Base Theorem.). This very useful theorem says log_{b}A = log_{c}A/log_{c}b Suppose, for example, you want log_{5}498. Our calculator doesn't have base 5, but the Crazy Base Theorem lets us calculate this by using base 10 or base e. The value is log(498)/log(5) = 3.858862792. If you raise 5 to the power 3.858862792, you will get 498. We could have used base e and and calculated ln(498)/ln(5). Try it! Let's prove the Crazy Base Theorem. (It certainly doesn't represent an obvious statement.)
Here's an example that makes use of the Crazy Base Theorem. If I want to know how long it will take money to double at a true rate of 6% a year, I must solve (1.06)^{x} = 2. OK, we have x = log_{1.06}2 = log(2)/log(1.06) = 11.89, or about 12 years. ====== Properties of logs (These can be summarized in four statements):
Problem example: Express log(AB^{2}/C^{3}) as a sum and/or difference of logs.
Problem example: Write 4logX + 6logY as a single logarithm.
Problem: Evaluate log_{3}27^{5}. Solution: (with communication): log_{3}27^{5} = 5(log_{3}27) = 5(3) = 15. Problem: Evaluate log_{13}987. Solution (with communication): Using the Crazy Base Theorem, we have log_{13}987 = log(987)/log(13) = 2.688. Problem: Write log x^{7 }- log y^{2} + log w^{1/2 }as a single logarithm. Solution (with communication): log x^{7 }- log y^{2} + log w^{1/2} = log(x^{7}/y^{2}) + log÷(w) = log[x^{7}÷(w)/y^{2}]. Problem: Solve for x: ln 15 + 2ln x = ln 135. Solution (with communication): ln 15 + 2ln x = ln 135 ==> ln 15 + ln x^{2} = ln 135 ==> ln x^{2} = ln 135 - ln 15 = ln(135/15) = ln 9 ==> x^{2} = 9 ==> x = 3 or x = -3. Since ln x is defined only for x > 0, the only solution to the original equation is x = 3. |