Exponents and Logarithms

Logarithmic Functions A logarithmic function is the inverse of an exponential function. For the function y = 2x, the inverse is x = 2y. In order to solve this inverse equation for y, we write it in logarithmic form. x = 2y is written as y = log2x and is read as “y = the logarithm of x to base 2”. y = 2x 1 2 4 8 16 1 2 4 8 16 y = log2x (x = 2y)

Graphing the Logarithmic Function y = x y = 2x y = log2x

Comparing Exponential and Logarithmic Function Graphs y = 2x The y-intercept is 1. y = x There is no x-intercept. The domain is {x | x Î R}. The range is {y | y > 0}. y = log2x There is a horizontal asymptote at y = 0. There is no y-intercept. The x-intercept is 1. The graph of y = 2x has been reflected in the line y = x, to give the graph y = log2x. The domain is {x | x > 0}. The range is {y | y Î R}. There is a vertical asymptote at x = 0.

Logarithm/Exponent Forms Consider 72 = 49. 2 is the exponent of the power, to which 7 is raised, to equal 49. The logarithm of 49 to the base 7 is equal to 2 (log749 = 2). Exponential notation Logarithmic form log749 = 2 72 = 49 In general: If bx = N, then logbN = x. State in logarithmic form: State in exponential form: a) 63 = 216 log6216 = 3 a) log5125 = 3 53 = 125 b) 42 = 16 log416 = 2 b) log2128= 7 27 = 128

Exponent to Logarithmic form State in logarithmic form: a) b) log2 32 = 3x + 2

Evaluating Logarithms Note: log2128 = log227 = 7 log327 = log333 = 3 log2128 = x 2x = 128 2x = 27 x = 7 log327 = x 3x = 27 3x = 33 x = 3 3. log556 = 6 logaam = m 4. log816 5. log81 log816 = x 8x = 16 23x = 24 3x = 4 log81 = x 8x = 1 8x = 80 x = 0 loga1 = 0

Evaluating Logarithms 6. log4(log338) 7. = x log48 = x 4x = 8 22x = 23 2x = 3 2x = 1 9. Given log165 = x, and log84 = y, express log220 in terms of x and y. 8. log165 = x log84 = y = 23 = 8 16x = 5 24x = 5 8y = 4 23y = 4 log220 = log2(4 x 5) = log2(23y x 24x) = log2(23y + 4x) = 3y + 4x

Evaluating Base 10 Logs Base 10 logarithms are called common logs. Using your calculator, evaluate to 3 decimal places: a) log1025 b) log100.32 c) log102 1.398 -0.495 0.301 Evaluate log29: log29 = x 2x = 9 Change of base formula: log 2x = log 9 xlog 2 = log 9 x = 3.170

Evaluating Logs Given log3a = 1.43 and log4b = 1.86, determine logba. log3a = 1.43 a = 31.43 log a = 1.43log 3 log4b = 1.86 b = 41.86 log b = 1.86 log 4 logba = 0.609