LOGARITHMS   The inverse of the exponential function-- Logarithms Below is the graph of          y  =  2x   and its inverse which we defined as         y   =  log2 x We see that the logarithm function y = logbx has the following properties: The x - intercept is (1,0). There is a vertical asymptote at x = 0. The domain is {x | x > 0}. The range is all real numbers. The graph goes through (b,1).   Evaluating Logarithms Example Evaluate          log3 81   Solution:   We run the "hook" as shown below and write          3y  =  81  so that          y  =  4    Exercises Evaluate the following   log10 100,000,000 log5 (1/125) log27 9 Example Solve         log9 x = 2   Solution  We write          x  =  92  =  81   Exercises Solve log2 x  =  5 logb 64  =  3 Inverse Properties of Logs Since logs and exponents cancel each other we have:         blogb x = x  and         logb bx = x   Example     2log2 3 = 3 and         log4 45 = 5     Three Properties of Logs         Property 1:  logb (uv)  = logb u  +  logb v    (The Product to Sum Rule)         Property 2:  logb (u/v) = logb u  -  logb v    (The Quotient to Difference Rule)          Property 3:  logb ur   =  rlogb u                    (The Power Rule)   Proof of the power rule We have the rule for exponents: Canceling the b we get         logb ur   = rlogb u Example Expand:           log2 (xy2/z) by property 2 we have:         log2 (xy2) -  log2 z by property 1 we have         log2 x + log2 y2  - log2 z By property 3 we have         log2 x + 2 log2 y - log2 z   Exercise Try to expand:  Example Write as a single logarithm:         4 log2 x - 1/2 log2 y  + log2 z Solution: We first use property 3 to write:         log2 x4 - log2 y1/2  + log2 z Now we use property 2:         log2 x4/y1/2  +  log2 z Finally, we use property 3:                    x4z          log2 (           )                     y1/2                      Exercise Write the following as a single logarithm:         1/3 log3 x + 2 log3 y  - 3 log3 z Example Suppose that           log2 3 = 1.58  and that          log2 5 = 2.32 Find          log2 90   Solution Since          90 = (2)(5)(32) We have         log2 90  =  log2 (2)(5)(32)  =  log2 2 + log2 5 + log2 32         =  1 + 2.32 + 2log2 3         =  1 + 2.32 + 2(1.58)  =  6.48   Exercise Find          log2 40/27 Back to the Intermediate Algebra (Math 154) Home Page