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Jim Lamb ers Math 1B Fall Quarter 2004- Le ture 4 Notes
These notes orresp ond to Se tions 2.6 and 4.3 in the text.
Logarithmi Fun tions
Inverse Fun tions
In Le ture 1, we brie y dis ussed the notion of an inverse fun tion. If f is a one-to-one fun tion, then it has an inverse fun tion, denoted by f ^1 , su h that
f ^1 (f (x)) = x and f (f ^1 (y )) = y ; (1)
for any x in the domain of f , and any y in the range of f. In words, these equations, whi h are known as the an el lation equation, show that f and f ^1 des rib e pro esses that undo one another. That is, if f is applied to an input x to pro du e an output y , and f ^1 is applied to y , then the result is the original input, x. Similarly, if f ^1 is applied to an input y to obtain an output x, and f is applied to x, then the result is y. The an ellation equations an b e used to obtain an equivalent hara terization, or de nition, of the inverse fun tion that is a tually more useful for the task of nding the inverse of a given fun tion f. We now state this de nition.
De nition 1 (Inverse Fun tion) Let f be a one-to-one fun tion with domain D and range R. The inverse fun tion of f , denoted by f ^1 , is the fun tion with domain R and range D de ned by the statement x = f ^1 (y ) if and only if y = f (x); (2)
where x belongs to D and y belongs to R.
The reason why this de nition of the inverse fun tion is useful for nding the inverse of a given fun tion f is that we an solve the equation y = f (x) for x, and then use this de nition to on lude that the expression we obtain for x is equal to f ^1 (y ), and therefore we will have found the inverse fun tion. The following result provides an easy way of determining whether a fun tion is one-to-one, and therefore would have an inverse. Re all that a fun tion f is in reasing if f (x) > f (y ) whenever x > y , and de reasing if f (x) < f (y ) whenever x > y.
Theorem 1 If a fun tion f is in reasing on its entire domain, or de reasing on its entire domain, then f is one-to-one.
Pro of Supp ose that f is in reasing. Then, if x < y , then f (x) < f (y ). Let x and y b e any two real numb ers in the domain of f su h that x 6 = y. Then, we must have x < y or x > y. If x < y , then f (x) < f (y ), and if x > y , then f (x) > f (y ). Therefore, f (x) 6 = f (y ). Sin e this is true for any x and y in the domain of f , we on lude that f is one-to-one. The pro of for the ase where f is de reasing is similar. 2
De nition of the Logarithmi Fun tion
In this le ture, we dis uss the inverse fun tion of the exp onential fun tion with base b, f (x) = bx^. As dis ussed in le ture 2, bx^ is an in reasing fun tion if b > 1, and it is a de reasing fun tion if 0 < b < 1. In either ase, by Theorem 1, bx^ is one-to-one. It follows that bx^ has an inverse fun tion. We now pre isely de ne this inverse fun tion.
De nition 2 (Logarithmi Fun tion) Let b > 0 , and assume that b 6 = 1. Let f (y ) = by^ be the exponential fun tion with base b. The logarithmi fun tion with base b is the inverse fun tion of f , and is denoted by f ^1 (x) = log (^) b x. That is,
y = log (^) b x if and only if by^ = x; (3)
where x > 0 and y is a real number. The statement y = log (^) b x in equation (3) is al led the logarithmi form, and the statement by^ = x in equation (3) is al led the exp onential form. The number y in equation (3) is al led the logarithm, or log, to base b of x.
It is imp ortant to realize that the value of log (^) b x is an exp onent. Sp e i ally, if y = log (^) b x, then y is the exp onent to whi h b must b e raised in order to obtain x. The de nition provides a very useful statement for solving equations involving exp onential and logarithmi fun tions: the equivalen e of the exp onential form by^ = x and the logarithmi form y = log (^) b x. It is imp ortant to b e able to onvert b etween these forms in order to solve su h equations.
Prop erties of Logarithmi Fun tions
The prop erties of exp onential fun tions intro du ed in Le ture 2 an b e used to derive similar prop erties for logarithmi fun tions. We will see that these prop erties are extremely useful in solving problems involving logarithmi fun tions. In the statement of the following theorem that lists these prop erties, we use the symb ol (), whi h means \if and only if." That is, the statement p () q means that the statements p and q are true under exa tly the same ir umstan es, so, in e e t, they are equivalent statements.
Theorem 2 (Properties of Logarithmi Fun tions) Let b, x, y , and r be real numbers, where b, x and y are positive, and b 6 = 1. Then
- Let s = log (^) b x and t = log (^) b y. Then, rewriting the de nitions of s and t in exp onential form, we obtain x = bs^ and y = bt^. Re all that for any real numb ers s and t, bs+t^ = bs^ bt^. This implies that bs+t^ = xy. If we rewrite this equation in logarithmi form, we obtain log (^) b xy = s + t. By the de nitions of s and t, we obtain log (^) b xy = log (^) b x + log (^) b y.
- Let s = log (^) b x and t = log (^) b y. Then, rewriting the de nitions of s and t in exp onential form, we obtain x = bs^ and y = bt^. Re all that for any real numb ers s and t, bs t^ = bs^ =bt^. This implies that bs t^ = x=y. If we rewrite this equation in logarithmi form, we obtain log (^) b x=y = s t. By the de nitions of s and t, we obtain log (^) b x=y = log (^) b x log (^) b y.
- Let s = log (^) b x. Then, rewriting the de nition of s in exp onential form, we obtain x = bs^. Re all that for any real numb ers r and s, bsr^ = (bs^ )r^. This implies that bsr^ = xr^. If we rewrite this equation in logarithmi form, we obtain log (^) b xr^ = sr. By the de nitions of s and t, we obtain log (^) b xr^ = r log (^) b x.
- Let s = log (^) b x and t = log (^) b y. Then, rewriting the de nitions of s and t in exp onential form, we obtain x = bs^ and y = bt^. Re all that for any real numb ers s and t, bs^ = bt^ if and only if s = t. This implies that x = y if and only if s = t. By the de nitions of s and t, we on lude that x = y if and only if log (^) b x = log (^) b y. This is logi ally equivalent to the statement that log (^) b x = log (^) b y if and only if x = y.
2 It is imp ortant to use aution when applying these prop erties. The following mistakes are very ommon: Confusing the log of a sum or di eren e with the sum or di eren e of logs. The expression log (^) b x + log (^) b y is a sum of logs, and it an b e simpli ed to log (^) b xy. The expression log (^) b (x + y ) is a log of a sum, and it annot b e simpli ed. Confusing the log of a produ t or quotient with the produ t or quotient of logs. The expression (log (^) b x)n^ , where n is a p ositive integer, is a pro du t of logs, and it annot b e simpli ed. The expression log (^) b xn^ is a log of a pro du t, and it an b e simpli ed to n log (^) b x. Similarly, the expression log (^) b x= log (^) b y is a quotient of logs, and it annot b e simpli ed. On the other hand, the expression log (^) b (x=y ) is a log of a quotient, and it an b e rewritten as log (^) b x log (^) b y.
Graphs of Logarithmi Fun tions
The graph of log (^) b x an b e obtained from the graph of bx^ by inter hanging the x- and y - o ordinates of ea h p oint on the graph of bx^ , sin e log (^) b x is its inverse fun tion. It follows that if b > 1, then log (^) b x is an in reasing fun tion, and if 0 < b < 1, then log (^) b x is a de reasing fun tion. Furthermore, the graph of every logarithmi fun tion passes through the p oint (1; 0), just as the graph of every exp onential fun tion passes through the p oint (0; 1). Finally, the graph of log (^) b x approa hes the verti al line x = 0, just as the graph of bx^ approa hes the horizontal line y = 0.
