The function e x and its inverse lnx
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y=3 x. y=e x. y=2 x. x. The function e x and its inverse, lnx. The functions like y = 2 x , y = 3 x are called exponential functions because the variable x is the power (exponent or index) of a base number. The graph of y = 2 x , y = 3 x and y = e x.

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The function e x and its inverse, lnx

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The function e x and its inverse lnx

y=3x

y=ex

y=2x

x

The function ex and its inverse, lnx

The functions like y = 2x, y = 3x are called exponential functions because the variable x is the power (exponent or index) of a base number.

The graph of y = 2x, y = 3x and y = ex

It is clear from the graph that the number e is somewhere between 2 and 3 but closer to 3 than 2.

This is a special number and its value correct to 8 decimal places is e =2.18281828.


Graph of e x and lnx

y

1

x

1

Graph of ex and lnx

y = ex

y = x

y = lnx

y = ex

y = lnx


Evaluating function e x and ln x

Evaluating function ex and lnx

Evaluate

(i) e2 (ii) e-3 (iii) ln0.5 (iv) ½ ln10

(i) 7.39

(ii) 0.0498

(iii) 7.39

(iv) 1.15

Find the value of x

(i) ln ex = 3 (ii) elnx = 5 (iii) e2lnx = 16 (iv) e-lnx = ½

(i) x = 3

(ii) x = 5

(iii) x = 4

(iv) x = 2


Solving equations involving e x and lnx

Solving equations involving ex and lnx

Solve for x

(i) 3e2x – 1 = 54

(ii) 3e2x –5ex = 2

 3e2x = 55

Let y = ex

 e2x = 18.333..

 3y2 – 5y = 2

 2x= ln18.333..

 3y2 – 5y – 2 = 0

 2x= 2.9087..

 (3y )(y )

 x= 1.45

 (3y + 1)(y - 2)

 y = - 1/3 or y = 2

 ex = 2  x = 0.693


Solving equations involving e x and lnx1

Solving equations involving ex and lnx

Solve for x

(i) ln(3x – 5) = 3.4

(ii) ln(3x + 1) – ln3 = 1

 ln((3x + 1)/3) = 1

 3x – 5 = e3.4

 ((3x + 1)/3) = e1

 3x – 5 = 29.964…

 ((3x + 1)/3) = 2.718…

 3x = 34.964…

 x= 11.7.

 3x + 1 = 8.1548..

 3x = 7.1548…

x = 2.38


Exponential decay growth

Exponential Decay/Growth

A quantity N is decreasing such that at time t

(a) Find the value of N when t = 5

(b) Find the value of t when t = 2

= 18.4

(a)

(b)

t = 16.1


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