Transcript 4.3 Powerpoint

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
ya
x
gives
xa
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.
log49 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)  log1 x
2
Graph the function
f ( x)  log1 x
2
Graph the function
f ( x)  log3 x
Graph the function
f ( x)  log3 x
Graph the function
f ( x)  log2 x  1
Graph the function
f ( x)  log2 x  1
Graph the function
f ( x)  log3 x  1
Graph the function
f ( x)  log3 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
log6 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
log5 8
Rewrite each expression. Assume all variables
represent positive real numbers, with
m nq
loga 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
logb 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.
log3 x  2  log3 x  log3 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 loga m  3 loga 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)
log3 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 log10 4
Assume that log102 = .3010
Find log10 5
7
log7 10
 10
log5 5  3
3
logr r
k 1
 k 1