Logarithmic Functions

y = log a x, is read “the logarithm, base a, of x,” or “log, base a, of x,” means “the exponent to which we raise a to get x.” Exponent Base Argument Is equivalent to

Example Solution a) log 3 81 b) log 3 1 c) log 3 (1/9) a) Think of log 3 81 as the exponent to which we raise 3 to get 81. That exponent is 4. Therefore, log 3 81 = 4. Simplify: b) We ask: “To what exponent do we raise 3 in order to get 1?” That exponent is 0. Thus, log 3 1 = 0. c) To what exponent do we raise 3 in order to get 1/9? Since 3 -2 = 1/9, we have log 3 (1/9) = –2.

Example Solution Simplify: Remember that log 5 23 is the exponent to which 5 is raised to get 23. Raising 5 to that exponent, we have It is important to remember that a logarithm is an exponent.

Example Graph y = f (x) = log 3 x. Solution y 1 3 1/3 9 1/ –1 2 –2 3

Common Logarithms Base-10 logarithms, called common logarithms, are useful because they have the same base as our “commonly” used decimal system, and it is one of two logarithms on our calculator. We’ll discuss this later. Example

Solution Example Graph: y = log (x/4) – 2 in the window [  2, 8] X [  5,5].

Equivalent Equations We use the definition of logarithm to rewrite a logarithmic equation as an equivalent exponential equation or the other way around: m = log a x is equivalent to a m = x.

Solution Example exponential equation: a) –m = log 3 x b) 6 = log a z Rewrite each as an equivalent a) –m = log 3 x is equivalent to 3  m = x b) 6 = log a z is equivalent to a 6 = z. The base remains the base. The logarithm is the exponent.

Solution Example logarithmic equation: a) 49 = 7 x b) x  2 = 9 Rewrite each as an equivalent a) 49 = 7 x is equivalent to x = log 7 49 b) x  2 = 9 is equivalent to –2 = log x 9. The base remains the base. The exponent is the logarithm.

Solving Certain Logarithmic Equations Logarithmic equations are often solved by rewriting them as equivalent exponential equations.

Example Solution Solve: a) log 3 x = –3; b) log x 4 = 2. a) log 3 x = –3 x = 3 –3 = 1/27 b) log x 4 = 2 4 = x 2 x = 2 or x = –2 Because all logarithmic bases must be positive, –2 cannot be a solution. The solution is 2. The solution is 1/27. The check is left to the student.

Solution Example Solve: a) log 6 36 = x; b) log 9 1 = t. a) log 6 36 = x 6 x = 36 x = 2 6 x = 6 2 b) log 9 1 = t 9 t = 1 9 t = 9 0 t = 0

= 0 = 1