Exercise 3.3.1
Recreate your sketch of the graph of y = 2 x on a square sheet of paper. Sketch the graph of
the inverse function by reversing the ordered pairs of points lying on the graph of y = 2 x. You can also see what the graph of the inverse will look like by holding the sheet of graph paper by its lower left and upper right corners and rotating 180o about the line y = x, so that you are looking at the back of the graph paper. Hold the paper up to a light source until you can see the graph through the paper. What you see will be the graph of the inverse.
Solution
Exercise 3.3.2
Recreate your sketch of the graph of y = ( 1 / 2 ) x on a square sheet of paper. Sketch the graph of
the inverse function.
Solution
Recall the method for finding the equation of an inverse function. First, replace the f ( x ) with y. Then exchange exchange each x with y and visa versa. Next, solve the new equation for y. Then replace that y with f -1( x ). Let us apply that method to the exponential function f ( x ) = b x.
Step 1: y = b x
Step 2: x = b y
Step 3: ? ? ? ?
How can we solve step 2 for y ? We want to find the value of y which, when it is the exponent of b, yields the value x. Such an exponent is called a logarithm. In this case, y would be the logarithm base b of x. That is, y = log b ( x ).
Thus,
Step 3: y = log b ( x )
Step 4: f -1( x ) = log b ( x )
The inverse of an exponential function is a logarithmic function.
This implies that the equation x = b y is equivalent to the equation y = log b ( x ).
This is a useful equivalence--remember it!
Exercise 3.3.3
Find the following logarithms:
(a) log 5 ( 25 )
(b) log 5 ( 5 )
(c) log 5 ( 1 / 5 )
Solution
Exercise 3.3.4
Sketch the graphs of the following logarithmic functions:
(a) f ( x ) = log 2 ( x )
(b) g ( x ) = log 0.5 ( x )
Solution
There are certain rules of logarithms which follow from the rules of exponents and the exponential-logarithmic equivalence mentioned above.
Recall, specifically, that
a m a n = a m + n
( a m ) n = a m n
Now, let m = log b ( p ) and n = log b ( q ). Then p = b m and q = b n.
Thus, pq = b mb n = b m + n = b ( log b ( p ) + log b ( q ) ).
So pq = b ( log b ( p ) + log b ( q ) ). Using the exponential-logarithmic equivalence to re-write this exponential statement as a logarithmic statement, we get the result
log b ( p ) + log b ( q ) = log b ( p q )
Exercise 3.3.5
Prove that log b ( p / q ) = log b ( p ) - log b ( q )
Solution
Let m = log b ( p r ). Then b m = p r, so b m / r = p. So log b ( p ) = m / r.
Thus, m = r log b ( p ) . Therefore
log b ( p r ) = r log b ( p )
Exercise 3.3.6
What logarithm rule derives from the exponential-logarithmic equivalence and the fact that
b 0 = 1 ? What rule derives from the fact that b 1 = b ?
Solution
Exercise 3.3.7
Given that log b ( 2 ) = 1.7095 and log b ( 3 ) = 2.7095, find the following:
(a) log b ( 6 )
(b) log b ( 72 )
(c) log b ( 1.5 )
(d) What is the value of b ?
Solution
Exercise 3.3.8
Use various rules of logarithms to rewrite the following in terms of log 7 ( x ), log 7 ( y ), and
log 7 ( z ):
Solution
Exercise 3.3.9
Use various rules of logarithms to rewrite the following as a single logarithm:
log 3 ( x ) + 2 log 3 ( y ) - (1 / 2 ) log 3 ( z )
Solution
There are two standard bases for logarithms, namely base 10, called the common base, and base e called the natural base.
When no base is specified, as in log ( x ), the common base 10 is assumed.
The natural base e is specified by writing ln ( x ) in place of log e ( x ).
Exercise 3.3.10
Find the following without using a calculator:
(a) log ( 0.01 )
(b) ln ( e 3 )
Solution
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