The Exponential and Log Functions

The Exponential and Log Functions

Introduction We are going to look at exponential functions
We will learn about a new ‘special’ number in Mathematics We will see how this number can be used in practical problems…

Teachings for Exercise 3A

Imagine you have £100 in a bank account Imagine your interest rate for the year is 100% You will receive 100% interest in one lump at the end of the year, so you will now have £200 in the bank However, you are offered a possible alternative way of being paid Your bank manager says, ‘If you like, you can have your 100% interest split into two 50% payments, one made halfway through the year, and one made at the end’ How much money will you have at the end of the year, doing it this way (and what would be the quickest calculation to work that out?) £100 x 1.52 = £225 Investigate further. What would happen if you split the interest into 4, or 10, or 100 smaller bits etc… 3A

Start Amount Payments Interest Each Payment Sum Total (2dp) £100 1 100% £100 x 2 £200 £100 2 50% £100 x 1.52 £225 £100 4 25% £100 x 1.254 £244.14 £100 8 12.5% £100 x £256.58 £100 10 10% £100 x 1.110 £259.37 £100 20 5% £100 x £265.33 £100 50 2% £100 x £269.16 £100 100 1% £100 x £270.48 £100 1,000 0.1% £100 x £271.69 £100 10,000 0.01% £100 x £271.81 £100 n 100/n £100 x (1 + 1/n)n £100e The larger the value of n, the better the accuracy of e… ( …) The value of e is irrational, like π… It also has another interesting property… 3A

y = x2 Gradient Functions You have already learnt about differentiation, that differentiating a graph function gives the gradient function… We can plot the gradient function on the graph itself… y = x2 So dy/dx = 2x y = 2x The Gradient at this point… … is this value here! 3A

y = 3x2 y = x3 Gradient Functions You have already learnt about differentiation, that differentiating a graph function gives the gradient function… We can plot the gradient function on the graph itself… y = x3 So dy/dx = 3x2 … is this value here! The Gradient at this point… And the Gradient is the same here! 3A

Gradient Functions y = 2x You have already learnt about differentiation, that differentiating a graph function gives the gradient function… We can plot the gradient function on the graph itself… y = 2x dy/dx = 2xln2 y = 2xln2 At this stage, you do not need to know where this comes from… 3A

y = 3xln3 Gradient Functions You have already learnt about differentiation, that differentiating a graph function gives the gradient function… We can plot the gradient function on the graph itself… y = 3x dy/dx = 3xln3 y = 3x At this stage, you do not need to know where this comes from… 3A

y = 2xln2 y = 3xln3 y = 3x y = 2x What has happened from the first graph to the second? The lines have crossed… Therefore the must be a value between 2 and 3 where the lines are equal… 3A

Gradient Functions y = ex If we plot a graph of ex, its gradient function is the same graph! This leads to an interesting conclusion… If y = ex Then dy/dx = ex as well! y = ex 3A

The Graph of ex is used rather than 2x or 3x because e is the increase when growth is continuous (ie growth is always happening, rather than in ‘chunks’) The majority of growth follows this format, bacterial growth is continuous for example. Transforming the graph can also allow graphs to show decreases, such as radioactive decay and depreciation 3A

y = 2ex y = ex You need to be able to sketch transformations of the graph y = ex y = ex y = 2ex (0,2) (0,1) f(x) 2f(x) (For the same set of inputs (x), the outputs (y) double) 3A

y = ex y = ex + 2 You need to be able to sketch transformations of the graph y = ex y = ex y = ex + 2 (0,3) (0,1) f(x) f(x) + 2 (For the same set of inputs (x), the outputs (y) increase by 2) 3A

y = ex You need to be able to sketch transformations of the graph y = ex y = ex y = -ex (0,1) f(x) (0,-1) y = -ex -f(x) (For the same set of inputs (x), the outputs (y) ‘swap signs’ 3A

y = e2x y = ex You need to be able to sketch transformations of the graph y = ex y = ex y = e2x (0,1) f(x) f(2x) (The same set of outputs (y) for half the inputs (x)) 3A

y = ex + 1 y = ex You need to be able to sketch transformations of the graph y = ex y = ex y = ex + 1 (0,e) (0,1) f(x) f(x + 1) (The same set of outputs (y) for inputs (x) one less than before…) We can work out the y-intercept by substituting in x = 0  This gives us e1 = e 3A

y = ex You need to be able to sketch transformations of the graph y = ex y = ex y = e-x y = e-x (0,1) (0,1) f(x) f(-x) (The same set of outputs (y) for inputs with the opposite sign… 3A

y = 10e-x y = ex You need to be able to sketch transformations of the graph y = ex Sketch the graph of: y = 10e-x y = e-x (0, 10) (0, 1) The graph of e-x, but with y values 10 times bigger… 3A

You need to be able to sketch transformations of the graph y = ex Sketch the graph of: y = 3 + 4e0.5x y = ex (0, 7) y = 3 + 4e0.5x y = e0.5x (0, 4) y = 4e0.5x (0, 1) The graph of e0.5x, but with y values 4 times bigger with 3 added on at the end… 3A

You need to be able to used the Exponential and Log Functions to solve problems… The Price of a used car is given by the formula: a) Calculate the value of the car when it is new  The new price implies t, the time, is 0…  Substitute t = 0 into the formula… P = 16000e - t 10 P = 16000e - 0 10 P = 16000e P = 16000e - t 10 P = £16000 3A

You need to be able to used the Exponential and Log Functions to solve problems… The Price of a used car is given by the formula: b) Calculate the value after 5 years…  5 years implies t = 5  Substitute t = 5 into the formula… P = 16000e - t 10 P = 16000e - 5 10 -0.5 P = 16000e P = 16000e - t 10 P = £ 3A

You need to be able to used the Exponential and Log Functions to solve problems… The Price of a used car is given by the formula: c) What is the implied value of the car in the long run (ie – what value does it tend towards?)  Imagine t tends towards infinity (gets really big) P = 16000e - t 10 P = x 0 P = £0 P = 16000e - t 10 1 (10√e)t e - t 10 Bigger t = Bigger denominator = Smaller Fraction value… 3A

You need to be able to used the Exponential and Log Functions to solve problems… The Price of a used car is given by the formula: d) Sketch the Graph of P against t  Value starts at £16000  Tends towards 0, but doesn’t get there… £16000 P = 16000e - t 10 P = 16000e - t 10 t  t is independent so goes on the x axis  P is dependant on t so goes on the y axis 3A

Teachings for Exercise 3B

You need to be able to solve equations involving natural logarithms and e  This is largely done in the same way as in C2 logarithms, but using ‘ln’ instead of ‘log’ Example Question 1 Take natural logs of both sides Use the ‘power’ law ln(e) = (e to the power something is e) Work out the answer or leave as a logarithm You do not necessarily need to write these steps… 3B

You need to be able to solve equations involving natural logarithms and e  This is largely done in the same way as in C2 logarithms, but using ‘ln’ instead of ‘log’ Example Question 1 Take natural logs Use the power law Subtract 2 Work out the answer or leave as a logarithm 3B

You need to be able to solve equations involving natural logarithms and e  This is largely done in the same way as in C2 logarithms, but using ‘ln’ instead of ‘log’ Example Question 1 ‘Reverse ln’ Work it out if needed… 3B

You need to be able to solve equations involving natural logarithms and e  This is largely done in the same way as in C2 logarithms, but using ‘ln’ instead of ‘log’ Example Question 1 ‘Reverse ln’ Add 2 Divide by 3 3B

y = ex You need to be able to plot and understand graphs of the function which is inverse to ex The inverse of ex is logex (usually written as lnx)  We know from chapter 2 that an inverse function is a reflection in the line y = x… y = x y = lnx (0,1) (1,0) 3B

y = 2lnx You need to be able to plot and understand graphs of the function which is inverse to ex y = lnx y = lnx f(x) (1,0) y = 2lnx 2f(x) All output (y) values doubled for the same input (x) values… 3B

y = lnx + 2 You need to be able to plot and understand graphs of the function which is inverse to ex y = lnx y = lnx f(x) (0.14,0) (1,0) y = lnx + 2 f(x) + 2 Let y = 0 All output (y) values increased by 2 for the same input (x) values… Subtract 2 Inverse ln Work out x! 3B

You need to be able to plot and understand graphs of the function which is inverse to ex y = lnx (1,0) y = lnx f(x) y = -lnx y = -lnx -f(x) All output (y) values ‘swap sign’ for the same input (x) values… 3B

You need to be able to plot and understand graphs of the function which is inverse to ex y = ln(2x) y = lnx (0.5,0) (1,0) y = ln(x) f(x) y = ln(2x) f(2x) All output (y) values the same, but for half the input (x) values… 3B

You need to be able to plot and understand graphs of the function which is inverse to ex y = ln(x + 2) y = lnx (0, ln2) (-1,0) (1,0) y = ln(x) f(x) y = ln(x + 2) f(x + 2) All output (y) values the same, but for input (x) values 2 less than before Let x = 0 Work it out (or leave as ln2) 3B

You need to be able to plot and understand graphs of the function which is inverse to ex y = ln(-x) y = lnx (-1,0) (1,0) y = ln(x) f(x) y = ln(-x) f(-x) All output (y) values the same, but for input (x) values with the opposite sign to before 3B

y = 3 + ln(2x) You need to be able to plot and understand graphs of the function which is inverse to ex y = ln(2x) y = lnx (0.025,0) Sketch the graph of: (1,0) y = 3 + ln(2x) Let y = 0 The graph of ln(2x), moved up 3 spaces… Subtract 3 Reverse ln Divide by 2 3B

You need to be able to plot and understand graphs of the function which is inverse to ex y = ln(3 + x) y = ln(3 - x) y = lnx (0,ln3) (2,0) Sketch the graph of: (-2,0) (1,0) y = ln(3 - x) The graph of ln(x), moved left 3 spaces, then reflected in the y axis. You must do the reflection last! Let x = 0 3B

You need to be able to plot and understand graphs of the function which is inverse to ex The number of elephants in a herd can be represented by the equation: Where n is the number of elephants and t is the time in years after 2003. Calculate the number of elephants in the herd in 2003  Implies t = 0 t = 0 e0 = 1 3B

You need to be able to plot and understand graphs of the function which is inverse to ex The number of elephants in a herd can be represented by the equation: Where n is the number of elephants and t is the time in years after 2003. b) Calculate the number of elephants in the herd in 2007  Implies t = 4 t = 4 Round to the nearest whole number 3B

You need to be able to plot and understand graphs of the function which is inverse to ex The number of elephants in a herd can be represented by the equation: Where n is the number of elephants and t is the time in years after 2003. Calculate the year when the population will first exceed 100 elephants  Implies N = 100 N = 100 Subtract 150 Divide by -80 Take natural logs Multiply by 40 = 2022 3B

You need to be able to plot and understand graphs of the function which is inverse to ex The number of elephants in a herd can be represented by the equation: Where n is the number of elephants and t is the time in years after 2003. d) What is the implied maximum number in the herd?  Implies t  ∞ Rearrange As t increases Denomintor becomes bigger Fraction becomes smaller, towards 0 3B

Summary We have learnt a new number, e, and seen what is stands for and where it is used We have plotted the graph of ex and lnx, which are inverse functions We have seen how to transform these graphs We have solved practical problems involving these We have also seen again how to solve logarithmic equations…

The Exponential and Log Functions

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