Chapter 4.3

Logarithms

LogarithmsThe previous section dealt with exponential function of the form y = ax for all positive values of a, where a ≠1.

LogarithmsThe horizontal line test shows that exponential functions are one-to-one,

and thus have inverse functions.

LogarithmsThe equation defining the inverse function is found by interchanging x and y in the equation that defines the function.

Doing so with

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.

yx axay gives

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 = ay.

Logarithmic EquationsThe 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.

327

8log x

Solve the equation.

2

5log4 x

Solve the equation.

x349 7log

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

xxf21log)(

Graph the function

xxf21log)(

Graph the function

xxf 3log)(

Graph the function

xxf 3log)(

Graph the function

1log)( 2 xxf

Graph the function

1log)( 2 xxf

Graph the function

1log)( 3 xxf

Graph the function

1log)( 3 xxf

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

97log6

Rewrite each expression. Assume all variables represent positive real numbers, with

7

15log9

Rewrite each expression. Assume all variables represent positive real numbers, with

8log5

Rewrite each expression. Assume all variables represent positive real numbers, with

2log

p

mnqa

Rewrite each expression. Assume all variables represent positive real numbers, with

3 2log ma

Rewrite each expression. Assume all variables represent positive real numbers, with

nmb z

yx 53

log

Write each expression as a single logarithm with coefficient 1. Assume all variables represent positive real numbers, with a ≠1 and b≠1.

2loglog2log 333 xx

Write each expression as a single logarithm with coefficient 1. Assume all variables represent positive real numbers, with a ≠1 and b≠1.

mm aa log3log2

Write each expression as a single logarithm with coefficient 1. Assume all variables represent positive real numbers, with a ≠1 and b≠1.

nmnm bbb2log2log

2

3log

2

1

There is no property of logarithms to rewrite a logarithm of a sum or difference. That is why, in Example 5 (a)

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.

2log3 x

Assume that log102 = .3010

4log Find 10

Assume that log102 = .3010

5log Find 10

107 10log7

35log 35

1log 1 kr kr