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e The exponential function A reminder from C2. Use your calculator (if you need to) to work out the following. Why are these the answers? (Hint: What does logax actually mean?) Question Answer Reason log39 = log5125 = log216 = log61 = log648 = log51/25 = 2 32 = 9 3 53 = 125 4 24 = 16 60 = 1 641/2 (= √64) = 8 -2 5-2 = 1/25 Follow-up question: Based on your answers to the above, what is an alternative way of writing the statement: y = logex ?

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**e What you have in fact just done is made x the subject of y = ln x.**

The exponential function What you have in fact just done is made x the subject of y = ln x. i.e. y = ln x x = ey If we now interchange y and x we obtain y = ex. What effect does this have on the graph of y = ln x?

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**e y = x y × (ln3, 3) × (ln2, 2) y = ln x × (3, ln3) × (0, 1)**

The exponential function y = x y × (ln3, 3) × (ln2, 2) y = ln x × (3, ln3) × (0, 1) × (2, ln2) × (1, 0) x Interchanging x and y has the effect of reflecting the graph in the line y = x.

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**e y = x y = ex y × (ln3, 3) × (ln2, 2) y = ln x × (3, ln3) × (0, 1)**

The exponential function y = ex y = x y × (ln3, 3) × (ln2, 2) y = ln x × (3, ln3) × (0, 1) × (2, ln2) × (1, 0) x The function y = ex is called the exponential function (more on this later).

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e The exponential function You know from your work on functions that reflecting in the line y = x gives an inverse function, so it follows that y = ex and y = ln x are the inverse of each other. What does this mean? Investigate, using your calculator, eln x and ln(ex) for different values of x. What is happening? These functions “undo” each other. ln(ex) = xlne (using the laws of logarithms) = x (using the fact that ln e = logee = 1. More on this later!

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e The exponential function Although the function ex is called the exponential function, in fact any function of the form ax is exponential. y = ex Use your graphical calculator to plot the following: y = 1.5x y = 2x y = ex y = 3x y = 1x y = 0.5x y = e-x y = (-2)x Discuss: Why are 6, 7 & 8 different to the rest? y y = 3x y = 2x y = 1.5x y = 1x = 1 y = 0.5x x

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e The exponential function The exponential function y = ex increases at an ever increasing rate. This is described as exponential growth. By contrast, the graph of y = e-x, shown below, approaches the x axis ever more slowly as x increases: this is called exponential decay. y This works because: y = e-x = (1/e)x. 1/e = 1/ = < 1. So, multiplying this by itself an increasing number of times will result in a smaller and smaller answer, similar to y = 0.5x. y = e-x x

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e The exponential function You will meet ex and ln x again later in this course where you will learn to differentiate and integrate them. In this section we will be focusing on practical applications which require you to use the ln button on your calculator.

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e The exponential function Example The number, N, of insects in a colony is given by N = 2000e0.1t where t is the number of days after observations have begun. Sketch the graph of N against t. What is the population of the colony after 20 days? How long does it take the colony to reach a population of ?

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e The exponential function Solution The number, N, of insects in a colony is given by N = 2000e0.1t where t is the number of days after observations have begun. Sketch the graph of N against t. N N = 2000e0.1t 2000 When t = 0, N = 2000e0 = 2000 O t

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e The exponential function Solution The number, N, of insects in a colony is given by N = 2000e0.1t where t is the number of days after observations have begun. What is the population of the colony after 20 days? When t = 20, N = 2000e0.1×20 = The population is insects.

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e The exponential function Solution The number, N, of insects in a colony is given by N = 2000e0.1t where t is the number of days after observations have begun. (iii) How long does it take the colony to reach a population of ? When N = , = 2000e0.1t 5 = e0.1t. Taking natural logarithms of both sides, ln 5 = ln(e0.1t) = 0.1t and so t = 10ln5 = 16.09… It takes just over 16 days for the population to reach Remember ln(ex) = x.

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e The exponential function Example The radioactive mass, M grams of a lump of material is given by M = 25e t where t is the time in seconds since the first observation. Sketch the graph of M against t. What is the initial size of the mass? What is the mass after 1 hour? The half-life of a radioactive substance is the time it takes to decay to half of its mass. What is the half-life of this material?

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e The exponential function Solution The radioactive mass, M grams of a lump of material is given by M = 25e t where t is the time in seconds since the first observation. Sketch the graph of M against t. M 25 O t

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e The exponential function Solution The radioactive mass, M grams of a lump of material is given by M = 25e t where t is the time in seconds since the first observation. What is the initial size of the mass? When t = 0, M = 25e0 = 25. The initial mass is 25g.

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e The exponential function Solution The radioactive mass, M grams of a lump of material is given by M = 25e t where t is the time in seconds since the first observation. (iii) What is the mass after 1 hour? After 1 hour, t = 3600. M = 25e ×3600. The mass after 1 hour is 0.33g (to 2 decimal places).

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e The exponential function Solution The radioactive mass, M grams of a lump of material is given by M = 25e t where t is the time in seconds since the first observation. (iv) The half-life of a radioactive substance is the time it takes to decay to half of its mass. What is the half-life of this material? The initial mass is 25g, so after one half-life, M = ½ × 25 = 12.5g. At this point the value of t is given by = 25e t. Dividing both sides by 25 gives = e t. Taking logarithms of both sides: ln 0.5 = ln e t = t. t = = (to 1 decimal place). The half-life is seconds. ln 0.5

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**e Example Make p the subject of ln(p) – ln(1 – p) = t.**

The exponential function Example Make p the subject of ln(p) – ln(1 – p) = t.

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**e Example Make p the subject of ln(p) – ln(1 – p) = t. Solution a b**

The exponential function Example Make p the subject of ln(p) – ln(1 – p) = t. Solution a b Using log a – log b = log p ln = t 1 - p Writing both sides as powers of e gives Remember elnx = x p e ln = et 1 - p p = et 1 - p p = et(1 – p) p = et – pet p + pet = et p(1 + et) = et et p = 1 + et

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e The exponential function Example Solve the equation e1-3x = 5.

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**e Example Solve the equation e1-3x = 5. Solution**

The exponential function Example Solve the equation e1-3x = 5. Solution Take natural logarithms of both sides: ln(e1-3x) = ln 5 Using the laws of logarithms: (1 – 3x)ln e = ln 5 Since ln e = 1: – 3x = ln 5 x = 1 – ln 5 = 3

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e The exponential function Example Solve the equation ln(2x + 1) = 3

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**e Example Solve the equation ln(2x + 1) = 3 Solution**

The exponential function Example Solve the equation ln(2x + 1) = 3 Solution ln(2x + 1) = 3 2x + 1 = e3 x = e3 – 1 = 9.54 2

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Chapter 4 – Logarithms The Questions in this revision are taken from the book so you will be able to find the answers in there.

Chapter 4 – Logarithms The Questions in this revision are taken from the book so you will be able to find the answers in there.

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