Exponentials and Logarithms: Properties and Solutions, Assignments of Calculus

Solutions to problems related to the properties of exponential and logarithmic functions. Topics covered include graphing, function composition, and logarithmic identities. Students are expected to have a basic understanding of exponential functions and calculus.

Typology: Assignments

Pre 2010

Uploaded on 08/19/2009

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Math 201
Exponentials and logarithms Solutions
The purpose of this worksheet is for you to work through some of the basic properties of the
exponential and logarithmic functions. Recall that an exponential function is of the form P(t) =
P0atwhere P0is the “initial quantity” and ais the “base”.
(1) Use you calculator to graph P(t)=2tand Q(t) = .5tand sketch these graphs.
In general if a > 1 the exponential function will increase, this is called “exponential growth”.
If 0 <a<1 then the function will decrease, this is called “exponential decay”.
(2) Recall that the inverse function to fis denoted f1and is defined by the property
f1(y) = xexactly when f(x) = y.
What does this tell us about (ff1)(y)? What about (f1f)(x)?
(ff1)(y) = f(f1(y)) = f(x) = y, So (ff1)(y) = y.
(f1f)(x) = f1(f(x)) = f1(y) = x, So (f1f)(x) = x).
(3) Logarithms are defined to be inverse functions to exponential functions. The most com-
monly used bases are eand 10, therefore we will only discuss the logarithms associated with
these bases. First of all, what is e? Find this button on your calculator and write down the
first five digits. (In Math 202 you will learn how your calculator does this.)
e= 2.71828
(4) The inverse function to f(x) = exis f1(x) = ln(x) and is read as “log base e or “natural
log”. Use the compositions in (2) to derive two different rules about how exand ln(x)
interact.
x= (ff1)(x) = f(f1(x)) = f(ln(x)) = eln(x), So eln(x)=x
x= (f1f)(x) = f1(f(x)) = f1(ex) = ln(ex), So ln(ex) = x
(5) Repeat the previous part for g(x) = 10xwith inverse g1(x) = log(x), that is “log base 10”.
10log(x)=xand log(10x) = x
(6) Use you calculator to graph ln(x) and log(x). Sketch the graphs.
(7) Use (4) to solve the equation 17 = 2(ex+1)
17
2=ex+1
(1)
ln 17
2= ln(ex+1)(2)
ln 17
2=x+ 1(3)
ln 17
21 = x(4)
1
pf2

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Math 201

Exponentials and logarithms Solutions

The purpose of this worksheet is for you to work through some of the basic properties of the exponential and logarithmic functions. Recall that an exponential function is of the form P (t) = P 0 at^ where P 0 is the “initial quantity” and a is the “base”.

(1) Use you calculator to graph P (t) = 2t^ and Q(t) =. 5 t^ and sketch these graphs. In general if a > 1 the exponential function will increase, this is called “exponential growth”. If 0 < a < 1 then the function will decrease, this is called “exponential decay”.

(2) Recall that the inverse function to f is denoted f −^1 and is defined by the property

f −^1 (y) = x exactly when f (x) = y.

What does this tell us about (f ◦ f −^1 )(y)? What about (f −^1 ◦ f )(x)? (f ◦ f −^1 )(y) = f (f −^1 (y)) = f (x) = y, So (f ◦ f −^1 )(y) = y. (f −^1 ◦ f )(x) = f −^1 (f (x)) = f −^1 (y) = x, So (f −^1 ◦ f )(x) = x).

(3) Logarithms are defined to be inverse functions to exponential functions. The most com- monly used bases are e and 10, therefore we will only discuss the logarithms associated with these bases. First of all, what is e? Find this button on your calculator and write down the first five digits. (In Math 202 you will learn how your calculator does this.) e = 2. 71828

(4) The inverse function to f (x) = ex^ is f −^1 (x) = ln(x) and is read as “log base e” or “natural log”. Use the compositions in (2) to derive two different rules about how ex^ and ln(x) interact. x = (f ◦ f −^1 )(x) = f (f −^1 (x)) = f (ln(x)) = eln(x), So eln(x)^ = x x = (f −^1 ◦ f )(x) = f −^1 (f (x)) = f −^1 (ex) = ln(ex), So ln(ex) = x

(5) Repeat the previous part for g(x) = 10x^ with inverse g−^1 (x) = log(x), that is “log base 10”. 10 log(x)^ = x and log(10x) = x

(6) Use you calculator to graph ln(x) and log(x). Sketch the graphs.

(7) Use (4) to solve the equation 17 = 2(ex+1)

(1) = ex+

ln

(2) = ln(ex+1)

ln

(3) = x + 1

ln

(4) − 1 = x

1

2

(8) Use (4) to rewrite P (t) = P 0 at^ as an exponential function with base e, that is P (t) = P 0 ekt, where k only depends on a. Since a = eln(a)^ we can write P (t) = P 0 at^ as P (t) = P 0 eln(a)t. (9) There are three other important properties of logarithms. (a) ln(AB) = ln(A) + ln(B) (b) ln

( A

B

= ln(A) − ln(B) (c) ln(Ak) = k ln(A) Just to demonstrate that these come from laws of exponents and (4) I will prove the first of these properties. eln(A)+ln(B)^ = eln(A)eln(B)^ = AB = eln(AB) The other properties follow from similar arguments.

(10) Use the previous parts to solve the following equation 200 = 10(3x).

(5) 20 = 3x

(6) ln(20) = ln(3x)

(7) ln(20) = x ln(3)

ln(20) ln(3)

(8) = x

(11) Assume that a > 1. How long will it take for the initial quantity in the function P (t) = P 0 at to double? If 0 < a < 1 how long will it take for there to be half of the initial quantity?

(9) 2 P 0 = P 0 at

(10) 2 = at

(11) ln(2) = ln(at)

(12) ln(2) = t ln(a)

ln(2) ln(a)

(13) = t

The same thing works for half-life, just replace the 2 by 12.