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Chapter 4.3 Logarithms Logarithms The previous section dealt with exponential function of the form y = ax for all positive values of a, where a ≠1. Logarithms The horizontal line test shows that exponential functions are one-to-one, and thus have inverse functions. Logarithms The equation defining the inverse function is found by interchanging x and y in the equation that defines the function. Doing so with ya x gives xa y As the equation of the inverse function of the exponential function defined by y = ax. This equation can be solved for y by using the following definition. The “log” in the definition above is an abbreviation for logarithm. Read logax as “the logarithm to the base a of x.” By the definition of logarighm, if y = logax, then the power to which a must be raised to obtain x is y, or x = a y. Logarithmic Equations The definition of logarithm can be used to solve a logarithmic equation, and equation with a logarithm in at least one term, by first writing the equation in exponential form. Solve the equation. 8 log x 3 27 Solve the equation. 5 log 4 x 2 Solve the equation. log 49 7 x 3 Exponential and logarithmic functions are inverses of each other. The graph of y = 2x is shown in red in the figure. The graph of the inverse is found by reflecting the graph of y = 2x across the line y = x. The graph of the inverse function, defined by y = log2x, shown in blue, has the y-axis as a vertical asymptote. Since the domain of an exponential function is th set of all real numbers, the range of a logarithmic function also will be the set of all real numbers. In the same way, both the range of an exponential function and the domainof a logarithmic function are the set of all positive real numbers, so logarithms can be found for positive numbers only. Graph the function f ( x) log 1 x 2 Graph the function f ( x) log 1 x 2 Graph the function f ( x) log 3 x Graph the function f ( x) log 3 x Graph the function f ( x) log 2 x 1 Graph the function f ( x) log 2 x 1 Graph the function f ( x) log 3 x 1 Graph the function f ( x) log 3 x 1 Since a logarithmic statement can be written as an exponential statement, it is not surprising that the properties of logarithms are based on the properties of exponents. The properties of exponents allow us to change the form of logarithmic statements so that products can be converted to sums, quotienst can be converted to differences, and powers can be converted to products. Two additional properties of logarithms follow directly from the definition of loga 1=0 and logaa = 1 Proof To prove the product property let m = logbx and n = logby logbx = m means bm = x logby = n means bn = y Now consider the product xy xy = bm bn xy = bm+n logbxy = m +n logbxy = logbx + logby Rewrite each expression. Assume all variables represent positive real numbers, with log 6 7 9 Rewrite each expression. Assume all variables represent positive real numbers, with 15 log 9 7 Rewrite each expression. Assume all variables represent positive real numbers, with log 5 8 Rewrite each expression. Assume all variables represent positive real numbers, with mnq log a 2 p Rewrite each expression. Assume all variables represent positive real numbers, with log a 3 m 2 Rewrite each expression. Assume all variables represent positive real numbers, with log b n x3 y 5 zm Write each expression as a single logarithm with coefficient 1. Assume all variables represent positive real numbers, with a ≠1 and b≠1. log 3 x 2 log 3 x log 3 2 Write each expression as a single logarithm with coefficient 1. Assume all variables represent positive real numbers, with a ≠1 and b≠1. 2 log a m 3 log a m Write each expression as a single logarithm with coefficient 1. Assume all variables represent positive real numbers, with a ≠1 and b≠1. 1 3 log b m log b 2n log b m 2 n 2 2 There is no property of logarithms to rewrite a logarithm of a sum or difference. That is why, in Example 5 (a) log 3 x 2 was not written as log3x + log32 The distributive property does not apply in a situation like this because log3(x+y) is one term; “log” is a function name, not a factor. Assume that log102 = .3010 Find log 10 4 Assume that log102 = .3010 Find log 10 5 7 log7 10 10 log 5 5 3 3 log r r k 1 k 1