# 1.5 Exponential and logarithmic functions  (Page 7/17)

 Page 7 / 17

## Evaluating hyperbolic functions

1. Simplify $\text{sinh}\left(5\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\right).$
2. If $\text{sinh}\phantom{\rule{0.1em}{0ex}}x=3\text{/}4,$ find the values of the remaining five hyperbolic functions.
1. Using the definition of the $\text{sinh}$ function, we write
$\text{sinh}\left(5\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\right)=\frac{{e}^{5\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x}-{e}^{-5\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x}}{2}=\frac{{e}^{\text{ln}\left({x}^{5}\right)}-{e}^{\text{ln}\left({x}^{-5}\right)}}{2}=\frac{{x}^{5}-{x}^{-5}}{2}.$
2. Using the identity ${\text{cosh}}^{2}x-{\text{sinh}}^{2}x=1,$ we see that
${\text{cosh}}^{2}x=1+{\left(\frac{3}{4}\right)}^{2}=\frac{25}{16}.$

Since $\text{cosh}\phantom{\rule{0.1em}{0ex}}x\ge 1$ for all $x,$ we must have $\text{cosh}\phantom{\rule{0.1em}{0ex}}x=5\text{/}4.$ Then, using the definitions for the other hyperbolic functions, we conclude that $\text{tanh}\phantom{\rule{0.1em}{0ex}}x=3\text{/}5,\text{csch}\phantom{\rule{0.1em}{0ex}}x=4\text{/}3,\text{sech}\phantom{\rule{0.1em}{0ex}}x=4\text{/}5,$ and $\text{coth}\phantom{\rule{0.1em}{0ex}}x=5\text{/}3.$

Simplify $\text{cosh}\left(2\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\right).$

$\left({x}^{2}+{x}^{-2}\right)\text{/}2$

## Inverse hyperbolic functions

From the graphs of the hyperbolic functions, we see that all of them are one-to-one except $\text{cosh}\phantom{\rule{0.1em}{0ex}}x$ and $\text{sech}\phantom{\rule{0.1em}{0ex}}x.$ If we restrict the domains of these two functions to the interval $\left[0,\infty \right),$ then all the hyperbolic functions are one-to-one, and we can define the inverse hyperbolic functions    . Since the hyperbolic functions themselves involve exponential functions, the inverse hyperbolic functions involve logarithmic functions.

## Definition

Inverse Hyperbolic Functions

$\begin{array}{cccc}{\text{sinh}}^{-1}x=\text{arcsinh}\phantom{\rule{0.1em}{0ex}}x=\text{ln}\left(x+\sqrt{{x}^{2}+1}\right)\hfill & & & {\text{cosh}}^{-1}x=\text{arccosh}\phantom{\rule{0.1em}{0ex}}x=\text{ln}\left(x+\sqrt{{x}^{2}-1}\right)\hfill \\ {\text{tanh}}^{-1}x=\text{arctanh}\phantom{\rule{0.1em}{0ex}}x=\frac{1}{2}\text{ln}\left(\frac{1+x}{1-x}\right)\hfill & & & {\text{coth}}^{-1}x=\text{arccot}\phantom{\rule{0.1em}{0ex}}x=\frac{1}{2}\text{ln}\left(\frac{x+1}{x-1}\right)\hfill \\ {\text{sech}}^{-1}x=\text{arcsech}\phantom{\rule{0.1em}{0ex}}x=\text{ln}\left(\frac{1+\sqrt{1-{x}^{2}}}{x}\right)\hfill & & & {\text{csch}}^{-1}x=\text{arccsch}\phantom{\rule{0.1em}{0ex}}x=\text{ln}\left(\frac{1}{x}+\frac{\sqrt{1+{x}^{2}}}{|x|}\right)\hfill \end{array}$

Let’s look at how to derive the first equation. The others follow similarly. Suppose $y={\text{sinh}}^{-1}x.$ Then, $x=\text{sinh}\phantom{\rule{0.1em}{0ex}}y$ and, by the definition of the hyperbolic sine function, $x=\frac{{e}^{y}-{e}^{\text{−}y}}{2}.$ Therefore,

${e}^{y}-2x-{e}^{\text{−}y}=0.$

Multiplying this equation by ${e}^{y},$ we obtain

${e}^{2y}-2x{e}^{y}-1=0.$

This can be solved like a quadratic equation, with the solution

${e}^{y}=\frac{2x±\sqrt{4{x}^{2}+4}}{2}=x±\sqrt{{x}^{2}+1}.$

Since ${e}^{y}>0,$ the only solution is the one with the positive sign. Applying the natural logarithm to both sides of the equation, we conclude that

$y=\text{ln}\left(x+\sqrt{{x}^{2}+1}\right).$

## Evaluating inverse hyperbolic functions

Evaluate each of the following expressions.

${\text{sinh}}^{-1}\left(2\right)$
${\text{tanh}}^{-1}\left(1\text{/}4\right)$

${\text{sinh}}^{-1}\left(2\right)=\text{ln}\left(2+\sqrt{{2}^{2}+1}\right)=\text{ln}\left(2+\sqrt{5}\right)\approx 1.4436$

${\text{tanh}}^{-1}\left(1\text{/}4\right)=\frac{1}{2}\text{ln}\left(\frac{1+1\text{/}4}{1-1\text{/}4}\right)=\frac{1}{2}\text{ln}\left(\frac{5\text{/}4}{3\text{/}4}\right)=\frac{1}{2}\text{ln}\left(\frac{5}{3}\right)\approx 0.2554$

Evaluate ${\text{tanh}}^{-1}\left(1\text{/}2\right).$

$\frac{1}{2}\text{ln}\left(3\right)\approx 0.5493.$

## Key concepts

• The exponential function $y={b}^{x}$ is increasing if $b>1$ and decreasing if $0 Its domain is $\left(\text{−}\infty ,\infty \right)$ and its range is $\left(0,\infty \right).$
• The logarithmic function $y={\text{log}}_{b}\left(x\right)$ is the inverse of $y={b}^{x}.$ Its domain is $\left(0,\infty \right)$ and its range is $\left(\text{−}\infty ,\infty \right).$
• The natural exponential function is $y={e}^{x}$ and the natural logarithmic function is $y=\text{ln}\phantom{\rule{0.1em}{0ex}}x={\text{log}}_{e}x.$
• Given an exponential function or logarithmic function in base $a,$ we can make a change of base to convert this function to any base $b>0,b\ne 1.$ We typically convert to base $e.$
• The hyperbolic functions involve combinations of the exponential functions ${e}^{x}$ and ${e}^{\text{−}x}.$ As a result, the inverse hyperbolic functions involve the natural logarithm.

For the following exercises, evaluate the given exponential functions as indicated, accurate to two significant digits after the decimal.

$f\left(x\right)={5}^{x}$ a. $x=3$ b. $x=\frac{1}{2}$ c. $x=\sqrt{2}$

a. 125 b. 2.24 c. 9.74

$f\left(x\right)={\left(0.3\right)}^{x}$ a. $x=-1$ b. $x=4$ c. $x=-1.5$

$f\left(x\right)={10}^{x}$ a. $x=-2$ b. $x=4$ c. $x=\frac{5}{3}$

a. 0.01 b. 10,000 c. 46.42

$f\left(x\right)={e}^{x}$ a. $x=2$ b. $x=-3.2$ c. $x=\pi$

For the following exercises, match the exponential equation to the correct graph.

1. $y={4}^{\text{−}x}$
2. $y={3}^{x-1}$
3. $y={2}^{x+1}$
4. $y={\left(\frac{1}{2}\right)}^{x}+4$
5. $y=\text{−}{3}^{\text{−}x}$
6. $y=1-{5}^{x}$

d

b

e

For the following exercises, sketch the graph of the exponential function. Determine the domain, range, and horizontal asymptote.

#### Questions & Answers

Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
mukul Reply
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base. a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec Pls help me solve
ade
show that lim f(x) + lim g(x)=m+l
BARNABAS Reply
list the basic elementary differentials
Chio Reply
Differentiation and integration
Okikiola Reply
yes
Damien
proper definition of derivative
Syed Reply
the maximum rate of change of one variable with respect to another variable
Amdad
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
Inembo Reply
what is calculus?
BISWAJIT Reply
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
Geoffrey Reply
what is x and how x=9.1 take?
Pravin Reply
what is f(x)
Inembo Reply
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
Adri
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
Adri
I dont understand what you wanna say by (A' n B^c)^c'
Adri
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Adri
Ok so the set is formed by vectors and not numbers
Adri
A vector of length n
Adri
But you can make a set out of matrixes as well
Adri
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
Wait what's your math level?
Adri
High-school?
Adri
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
Adri
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
Adri
I would say 24
Adri
Offer both
Adri
Sorry 20
Adri
Actually you have 40 - 4 =36 who offer maths or physics or both.
Adri
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
Adri
56-36=20 who give both courses... I would say that
Adri
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie
how do i evaluate integral of x^1/2 In x
ayo Reply
first you simplify the given expression, which gives (x^2/2). Then you now integrate the above simplified expression which finally gives( lnx^2).
Ahmad
by using integration product formula
Roha
find derivative f(x)=1/x
Mul Reply
-1/x^2, use the chain rule
Andrew
f(x)=x^3-2x
Mul
what is domin in this question
noman
all real numbers . except zero
Roha
please try to guide me how?
Meher
what do u want to ask
Roha
?
Roha
the domain of the function is all real number excluding zero, because the rational function 1/x is a representation of a fractional equation (precisely inverse function). As in elementary mathematics the concept of dividing by zero is nonexistence, so zero will not make the fractional statement
Mckenzie
a function's answer/range should not be in the form of 1/0 and there should be no imaginary no. say square root of any negative no. (-1)^1/2
Roha
domain means everywhere along the x axis. since this function is not discontinuous anywhere along the x axis, then the domain is said to be all values of x.
Andrew
Derivative of a function
Waqar
right andrew ... this function is only discontinuous at 0
Roha
of sorry, I didn't realize he was taking about the function 1/x ...I thought he was referring to the function x^3-2x.
Andrew
yep...it's 1/x...!!!
Roha
true and cannot be apart of the domain that makes up the relation of the graph y = 1/x. The value of the denominator of the rational function can never be zero, because the result of the output value (range value of the graph when x =0) is undefined.
Mckenzie
👍
Roha
Therefore, when x = 0 the image of the rational function does not exist at this domain value, but exist at all other x values (domain) that makes the equation functional, and the graph drawable.
Mckenzie
👍
Roha
Roha are u A Student
Lutf
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Roha
What is the first fundermental theory of Calculus?
ZIMBA Reply
do u mean fundamental theorem ?
Roha
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Adaeze Reply
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