Problem

Find the numerical value of each expression. (a)…

00:37

Question

Answered step-by-step

Problem 3 Easy Difficulty

Find the numerical value of each expression.
(a) $ \cosh (\ln 5) $
(b) $ \cosh 5 $

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Leon Druch

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Watch More Solved Questions in Chapter 3

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Video Transcript

in this problem, we want to determine what is the hyperbolic co sign Of the natural log of five. And what is the hyperbolic co sign of five. The hyperbolic cosine function is to find like this hyperbolic cosine of X is equal to E. To the X. Power plus E. To the negative X power all divided by two. So if we're going to find a hyperbolic coast sign of the log of five, we're simply going to substitute the log of five in everywhere you see X. So if the hyperbolic co sign of X equals E. T. D. X plus each and negative X over two. Then the hyperbolic co sign of the natural log of five. Okay, we're going to substitute lage five in for X. So we're going to have E. T. D. X. Which will be E. To the log of five plus E. To the negative X. So E to the negative log of five. All divided by two. Now, each of the negative log of five. When you have a negative exponent, that can be rewritten as a fraction where one is in the numerator and uh Then E to the positive Lage five would be in the denominator. So just like X to the -2 equals one over excited positive too. E. To the negative log of five equals one over E. Uh to the positive block of five. Now a lot of this is gonna simplify pretty nicely. Even though right now it might not look like it. E. And the natural log rhythm or inverse functions of each other. Uh so E to the log of five is simply five plus. Now we have one over E to the log of five. Once again E. And the natural algorithm or in verses of each other. So E to the natural log of five is simply just five. And so this green expression becomes 1/5. So we have five plus 1/5. All over. True. Well five plus 1/5 is five and 1/5 And we're dividing it by two. Let's change this up here to a an improper fraction. So this would be 26/5 26 5th divided by two is the same thing as 26 50 times one half. So we get 26/10 or 13 5th. So there is a the numerical answer for the hyperbolic co sign of the natural log of five. Okay, give us a little space to work here. Now we have to determine what is the hyperbolic co sign of five. Hyperbolic cosine of X. Is each of the X plus each negative X over two. So the hyperbolic co sign of five will simply be E to the fifth plus E. To the negative five. All divided by two. We want numerical answer. So we're simply going to put this into the calculator, putting each of the 50 plus each of negative fifth. All divided by two. In the calculator we get 74 Point to one. If we round to the nearest 100

Get More Help with this Textbook

James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem