The Laws of Logarithms - Lecture Notes | MATH 1330, Study notes of Pre-Calculus

Download The Laws of Logarithms - Lecture Notes | MATH 1330 and more Study notes Pre-Calculus in PDF only on Docsity! lSection3.3.doc 1 Math 1330 Section 3.3 Laws of Logarithms The Laws of Logarithms Let 0 and 1a a> ≠ . Let , ,A B C be real numbers with 0 and 0A B> > 1. log ( ) log ( ) log ( )a a aAB A B= + 2. log log ( ) log ( )a a a A A B B   = −    3. log ( ) log ( )Ca aA C A= 4. log ( )Ca a C= 5. log ( ) log ( ) if and only if a aA B A B= = The Change of Base Formula Let ,a b be positive real numbers that are not equal to 1. Let 0x > . Then log ( ) log ( ) log ( ) a b a x x b = These are the basic rules we need to work with logarithmic equations and simplify logarithmic expressions. All 5 of the laws and the change of base formula are derived directly from the definition of a logarithm given in Section 2.1. How? Let’s look at them one at a time: First, let’s call log ( )a A w= and log ( )a B y= , and then write these expressions in their exponential form wa A= and ya B= . 1. In exponential form *w y w yAB a a a += = , The logarithmic version of this equation is log ( )a AB w y= + . Then, substitute for w and y and you get log ( ) log ( ) log ( )a a aAB A B= + 2. In exponential form, w w y y A a a B a −= = . The logarithmic form of this equation is loga A w y B   = −    . Substituting back in for w and y shows log log ( ) log ( )a a a A A B B   = −    3. Remember that log ( )a A w= translates to wa A= . Raising both sides of this exponential equation to the C power gives ( )w C Ca A= . Then, remembering that raising to powers corresponds to multiplying exponents, this says *C w Ca A= . The logarithmic form of this equation is log ( ) * log ( )Ca aA C w C A= = . lSection3.3.doc 2 4. This law is a combination of the third law and the fact that log ( ) 1a a = . 5. This law is a statement of the fact that the logarithmic functions and exponential functions are one-to-one. 6. The change of base formula: start with log ( )b x k= and put into exponential form as kb x= . Then, using Law 5 and base a, we can take the logarithm of both sides in base a: log ( ) log ( )ka ab x= . Then, use law 3 to obtain ( ) ( )log loga ak b x= . Solving for k and then substituting for k gives the formula. Example 1: Rewrite each of the following expressions in a form that has no logarithms of a product, quotient, or power. A. 23log [ ( 1)]x x − B. 45log ( 7 )x C. 5 3 4( 1) ln 7 x x x  −  −  Example 2: Evaluate each of the following expressions using the laws of logarithms. A. 3 3log (15) log (5)− B. 3log( 0.1)