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…

The Exponential and Log Functions 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  = £225 Investigate further. What would happen if you split the interest into 4, or 10, or 100 smaller bits etc…

The Exponential and Log Functions 3A £100e£100 x (1 + 1 / n ) n100 / n n£100 £271.81£100 x %10,000£100 £271.69£100 x %1,000 £100 £270.48£100 x %100£100 £269.16£100 x %50£100 £265.33£100 x %20£100 £259.37£100 x %10£100 £256.58£100 x %8£100 £244.14£100 x %4£100 £225£100 x %2£100 £200£100 x 2100%1£100 Total (2dp)Sum Interest Each Payment Payments Start Amount 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… ( …)

The Exponential and Log Functions 3A 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 = x 2 So dy / dx = 2x y = x 2 y = 2x The Gradient at this point… … is this value here!

The Exponential and Log Functions 3A 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 = x 3 So dy / dx = 3x 2 y = x 3 y = 3x 2 The Gradient at this point… … is this value here! And the Gradient is the same here!

The Exponential and Log Functions 3A 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 = 2 x dy / dx = 2 x ln2 At this stage, you do not need to know where this comes from… y = 2 x y = 2 x ln2

The Exponential and Log Functions 3A 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 = 3 x dy / dx = 3 x ln3 At this stage, you do not need to know where this comes from… y = 3 x y = 3 x ln3

The Exponential and Log Functions 3A y = 3 x y = 3 x ln3 y = 2 x y = 2 x ln2 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…

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

The Exponential and Log Functions The Graph of e x is used rather than 2 x or 3 x 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

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

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

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

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

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

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

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

You need to be able to sketch transformations of the graph y = e x Sketch the graph of: y = 3 + 4e 0.5x 3A y = e x The graph of e 0.5x, but with y values 4 times bigger with 3 added on at the end… (0, 1) (0, 7) y = e 0.5x y = 4e 0.5x y = 3 + 4e 0.5x (0, 4) The Exponential and Log Functions

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… 3A P = 16000e - t 10 P = 16000e - t 10 P = 16000e P = 16000e 0 P = £16000

The Exponential and Log Functions 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… 3A P = 16000e - t 10 P = 16000e - t 10 P = 16000e P = 16000e -0.5 P = £

The Exponential and Log Functions 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) 3A P = 16000e - t 10 P = 16000e - t 10 P = x 0 P = £0 e - t 10 1 ( 10 √e) t Bigger t = Bigger denominator = Smaller Fraction value…

The Exponential and Log Functions 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… 3A P = 16000e - t 10 P = 16000e - t 10 P t £16000  t is independent so goes on the x axis  P is dependant on t so goes on the y axis

The Exponential and Log Functions 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’ 3B Example Question 1 Take natural logs of both sides Use the ‘power’ law ln(e) = 1 (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…

The Exponential and Log Functions 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’ 3B Example Question 1 Take natural logs Use the power law Subtract 2 Work out the answer or leave as a logarithm

The Exponential and Log Functions 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’ 3B Example Question 1 ‘Reverse ln’ Work it out if needed…

The Exponential and Log Functions 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’ 3B Example Question 1 ‘Reverse ln’ Add 2 Divide by 3

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

The Exponential and Log Functions You need to be able to plot and understand graphs of the function which is inverse to e x 3B y = lnx (1,0) y = lnxf(x) y = 2lnx2f(x) y = 2lnx All output (y) values doubled for the same input (x) values…

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

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

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

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

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

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

The Exponential and Log Functions You need to be able to plot and understand graphs of the function which is inverse to e x 3B y = lnx (1,0) Sketch the graph of: 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! y = ln(3 + x) (-2,0) (2,0) y = ln(3 - x) Let x = 0 (0,ln3) x = 3

The Exponential and Log Functions You need to be able to plot and understand graphs of the function which is inverse to e x 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 a)Calculate the number of elephants in the herd in 2003  Implies t = 0 3B t = 0 e 0 = 1

The Exponential and Log Functions You need to be able to plot and understand graphs of the function which is inverse to e x 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 b)Calculate the number of elephants in the herd in 2007  Implies t = 4 3B t = 4 Round to the nearest whole number

The Exponential and Log Functions You need to be able to plot and understand graphs of the function which is inverse to e x 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 c)Calculate the year when the population will first exceed 100 elephants  Implies N = 100 3B N = = 2022 Subtract 150 Divide by -80 Take natural logs Multiply by 40

The Exponential and Log Functions You need to be able to plot and understand graphs of the function which is inverse to e x 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 d)What is the implied maximum number in the herd?  Implies t  ∞ 3B Rearrange As t increases  Denomintor becomes bigger  Fraction becomes smaller, towards 0

The Exponential and Log Functions You need to be able to plot and understand graphs of the function which is inverse to e x f(x) = x – 1 g(x) = e x + 1 a) Calculate gf(4) 3B

The Exponential and Log Functions You need to be able to plot and understand graphs of the function which is inverse to e x f(x) = x – 1 g(x) = e x + 1 b) Calculate g -1 (x) 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 e x and lnx, which are inverse functions We have seen how to transform these graphs We have solved practical problems involving theese We have also seen again how to solve logarithmic equations…