Logarithmic and exponential functions

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Functions

Exponential and logarithmic functions

 Exponential Functions  Logarithmic Functions


Exponential Functions

 The function represented by f(x)= ex

is called an exponential function with base e and exponent x.  The domain of f is the set of all real numbers

which is ] – , [


Laws of Exponents ď ľ Lets start by these rules:


Limits


Derivative ď ľ Derivate of an exponential function:


Table of Variation  Derivate of an exponential function:  f(x)= ex then f’(x)= ex>0  f’(x)>0 then f(x) is strictly increasing


Graph ď ľ Sketch the graph of the exponential function f(x) = ex.

Solution First, recall that the domain of this function is the set of real numbers Next, putting x = 0 gives y = e0 = 1, which is the y-intercept. Y- intercept means the point where the graph cuts y-axis. So here we have the point (0,1) would be the intersection between the curve and y-axis. There is no x-intercept, there is no value of x for which y = 0, here we have to note that ex >0. This means ex is always positive so it is always above x-axis.


Graph  From the limit of ex at -∞ which is zero, we conclude that

there is a horizontal asymptote at y = 0 (x-axis).  Furthermore, ex increases without bound when x increases since f’(x)= ex >0  From the limit of ex at +∞ which is +∞, we conclude the range of f is the interval ]0, [.


Graph  Sketch the graph of the exponential function f(x) = ex.

Solution

y 4

f (x ) = e x

e

–e

e

x


Summary of Exponential Functions  The exponential function y = ex has the following

properties: 1. Its domain is ]– , [. 2. Its range is ]0, [. 3. Its graph passes through the point (0, 1) 4. It is continuous on ]– , [. 5. It is increasing on ]– , [


Exercise  Sketch the graph of the exponential function f(x) = e–x.

Solution: Please follow the same steps, you should get the below curve.  Sketching the graph: y 5

3

1 –3

–1

f(x) = e–x 1

3

x


Could you do it?  If you could perform all steps and get a correct curve, you

can proceed to logarithmic functions.  If not, please review the lesson.  If you need help, please contact us and we will provide more

resources and explanation to make it easier for you: ibrahim.dhaini.2014@gmail.com


Logarithmic Functions


Logarithms  We’ve discussed exponential equations of the form

y = ex  If we want to solve the above equation to get the value of x,

then we are searching for the logarithm of y.

✦ Logarithm of x to the base e y = ln (x) if and only if x = ey

(x > 0)


Examples ď ľ Solve ln x = 4 :

Solution ď ľ By definition, ln x = 4 implies x = e4 .


Logarithmic Notation

log x = log10 x ln x = loge x

Common logarithm Natural logarithm


Laws of Logarithms


Laws of Logarithms


Logarithmic Function ď ľ The function defined by f(x)= ln x

is called the logarithmic function. ď ľ The domain of f is the set of all positive numbers, that means we are allowed to use only positive values of x


Properties of Logarithmic Functions

 The logarithmic function

y = ln x has the following properties: 1. Its domain is ]0, [. 2. Its range is ]– , [. 3. Its graph passes through the point (1, 0). 4. It is continuous on ]0, [. 5. It is increasing on ]0, [


Graph  Sketch the graph of the function y = ln x.

Solution  We first sketch the graph of y = ex.  The required graph is the mirror image of the y graph of y = ex with respect to the line y = x:

y = ex

y=x

y = ln x 1 1

x


Graph  Lets use the properties of f(x)= ln x to sketch the graph.  Allowed values of x are from 0 to +∞ (x>0)  This function is strictly increasing ( f’(x) = 1/x >0)  Ln 1 = 0  Limit at 0 is -∞ so x=0 is a vertical asymptote  Limit at +∞ is +∞


Graph


Properties Relating Exponential and Logarithmic Functions

ď ľ Properties relating ex and ln x:

eln x = x ln ex = x

(x > 0) (for any real number x)


Equations with ln x  Solve the equation 5 ln x + 3 = 0.

Solution  Add – 3 to both sides of the equation and then divide both sides of the equation by 5 to obtain: 5ln x = -3 3 ln x = - = -0.6 5 and so: eln x = e -0.6 x = e -0.6 x » 0.55


Examples ď ľ Expand and simplify the expression:

x2 x2 - 1 x 2 ( x 2 - 1)1/2 ln = ln x e ex = ln x 2 + ln( x 2 - 1)1/2 - ln e x 1 = 2 ln x + ln( x 2 - 1) - x ln e 2 1 = 2 ln x + ln( x 2 - 1) - x 2


Still need some exercises? ď ľ In the attached presentation, you can find some exercises

to help you solve an equation with ln(x) and ex . Microsoft PowerPoint 97-2003 Presentation

Microsoft PowerPoint 97-2003 Presentation


End of ChaptE r


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