C3 Chapter 3: Exponential and Log Functions Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com Last modified: 26th October 2015

-2 -1 1 2 3 4 5 6 7 8 6 4 2 -2 -4 -6 𝑦= 2 𝑥 x -2 -1 1 2 3 y 0.25 0.5 4 8 ? ? ? ? ? ? This is known as an exponential function. It is useful for modelling things like: population growth/savings with compound interest. The key property of exponential growth is that: the output gets multiplied by some constant each time the input increases (by a unit). e.g. A rabbit population might get 40% larger each year. This is in contrast to linear growth where we add some constant each time. ? ? Get students to sketch axes and tables in their books. Click to Brosketch

Gradients of Exponential Functions You won’t yet be able to differentiate exponential functions 𝑦= 𝑎 𝑥 till C4. But I’ve calculated some gradients for you – click the black arrow to reveal the graph and gradient function. Function Gradient 𝑦= 1 𝑥 𝑑𝑦 𝑑𝑥 =0 > 𝑦= 1.5 𝑥 𝑑𝑦 𝑑𝑥 =0.41× 1.5 𝑥 > 𝑦= 2 𝑥 𝑑𝑦 𝑑𝑥 =0.69× 2 𝑥 > 𝑦= 2.5 𝑥 𝑑𝑦 𝑑𝑥 =0.92× 2.5 𝑥 > 𝑦= 3 𝑥 𝑑𝑦 𝑑𝑥 =1.10× 3 𝑥 > 𝑦= 3.5 𝑥 𝑑𝑦 𝑑𝑥 =1.25× 3.5 𝑥 > Can you estimate the base of the exponential function where the gradient function is the same as the function itself?

“The” Exponential Function Gradient 𝑒=2.71828… is known as Euler’s Constant. It is one of the five most fundamental constants in mathematics (0, 1, 𝑖, 𝑒, 𝜋) and we will explore it in the coming slides. But for the purposes of this chapter, the most important thing to appreciate is that: 𝑦= 𝑒 𝑥 → 𝑑𝑦 𝑑𝑥 = 𝑒 𝑥 Although any function of the form 𝑦= 𝑎 𝑥 is known as an exponential function, 𝑒 𝑥 is known as “the” exponential function. Yeah, it’s that big a deal… You can find the exponential function on your calculator, to the right (above the “ln” key) 𝑦= 1 𝑥 𝑑𝑦 𝑑𝑥 =0 > 𝑦= 1.5 𝑥 𝑑𝑦 𝑑𝑥 =0.41× 1.5 𝑥 > 𝑦= 2 𝑥 𝑑𝑦 𝑑𝑥 =0.69× 2 𝑥 > 𝑦= 2.5 𝑥 𝑑𝑦 𝑑𝑥 =0.92× 2.5 𝑥 > 𝑦= 𝑒 𝑥 𝑦= 3 𝑥 𝑑𝑦 𝑑𝑥 =1.10× 3 𝑥 > 𝑦= 3.5 𝑥 𝑑𝑦 𝑑𝑥 =1.25× 3.5 𝑥 𝑑𝑦 𝑑𝑥 = 𝑒 𝑥 >

Bernoulli’s Compound Interest Problem (This won’t be examined) You have £1. If you put it in a bank account with 100% interest, how much do you have a year later? What if the interest is split into 2 instalments of 50% interest, how much will I have? What about 3 instalments of 33.3%? And so on… No. Instalments Money at Maturity 1 2 1 =£2 2 1.5 2 =£2.25 3 1. 3 3 =£2.37 4 1.25 4 =£2.44 𝑛 1+ 1 𝑛 𝑛 ? ? ? ? Bernoulli is credited as first finding the value of ‘𝑒’, as the solution to this problem. Euler introduced the letter 𝑒 to represent the value for the base of a logarithm (which we’ll see). ? As 𝑛 becomes large, the money at maturity approaches £𝒆 𝑒= lim 𝑛→∞ 1+ 1 𝑛 𝑛

Examples ? ? ? ? ? ? ? 1 Sketch graphs of: 𝑦= 1 2 +4 𝑒 1 2 𝑥 𝑦= 𝑒 2𝑥 𝑦= 1 2 +4 𝑒 1 2 𝑥 𝑦= 𝑒 2𝑥 𝑦= 𝑒 −𝑥 𝑦 𝑦 𝑦 ? ? ? 𝑦= 𝑒 𝑥 4.5 𝑦= 1 2 1 1 𝑥 𝑥 𝑥 The 1 2 in the power doesn’t do anything terribly exciting except to make the line grow slower (as 𝑒 1 2 𝑥 = 𝑒 𝑥 ) 𝑒 2𝑥 = 𝑒 𝑥 2 , so we’re just squaring the 𝑦 values. Note that squaring values less than 1 make them smaller, so the line is below on the left. 𝑒 −𝑥 = 1 𝑒 𝑥 . So start with graph of 𝑦= 𝑒 𝑥 and reciprocate 𝑦 values. This is known as exponential decay. We have decay whenever the 𝑎<1 in 𝑎 𝑥 . 2 The price of a used car can be represented by the formula: 𝑃=1 000+16 000 𝑒 − 𝑡 10 Where 𝑃 is the price in £s and 𝑡 is the age in years. Calculate: The new price: 𝑷=𝟏𝟎𝟎𝟎+𝟏𝟔𝟎𝟎𝟎 𝒆 𝟎 =£𝟏𝟕𝟎𝟎𝟎 The value after 5 years: 𝑷=𝟏𝟎𝟎𝟎+𝟏𝟔𝟎𝟎𝟎 𝒆 − 𝟏 𝟐 =£𝟏𝟎𝟕𝟎𝟒.𝟒𝟗 The long term eventual value of the car? As 𝒕→∞, 𝒆 − 𝒕 𝟏𝟎 = 𝟏 𝒆 𝒕 𝟏𝟎 will tend to 0. Thus 𝑷→£𝟏𝟎𝟎𝟎 d) Sketch the function. ? 17000 ? 1000 ? ?

Test Your Understanding The population of Tiffin Boys since 1970 is modelled using the equation 𝑃=1200−400 𝑒 −0.1𝑥 Where 𝑥 is the number of years since 1970 and 𝑃 is the number of boys. How boys were there in: 1970 𝑷=𝟏𝟐𝟎𝟎−𝟒𝟎𝟎 𝒆 𝟎 =𝟏𝟐𝟎𝟎−𝟒𝟎𝟎=𝟖𝟎𝟎 2014 𝑷=𝟏𝟐𝟎𝟎−𝟒𝟎𝟎 𝒆 −𝟎.𝟏×𝟒𝟒 =𝟏𝟏𝟗𝟓 Far into the future 𝑷=𝟏𝟐𝟎𝟎 Sketch a graph to represent the number of Tiffin Boys over time. ? ? ? ? 𝑃 1200 800 𝑥

4 4 4 4 4 4 Inverse of Exponentials 𝑥 5 5√x 3 𝑥 log3 x 𝑒 𝑥 loge x 1024 In C2, we learnt what the inverse is of an exponential function. 1024 4 4 𝑥 5 ? 5√x 81 4 4 3 𝑥 ? log3 x 54.59 4 4 𝑒 𝑥 loge x ? ! log 𝑒 𝑥 is the natural log of 𝒙 and is written 𝐥𝐧 𝒙 (or in ‘proper maths’, just simply log 𝑥 . Although confusingly on a calculator, log 𝑥 means base 10, not base 𝑒)

Solving Equations ? ? ? ? ? ? Solve the following: E1 𝑒 𝑥 =3 𝒙= 𝐥𝐧 𝟑 2 𝑒 4𝑥 =3 𝒆 𝟒𝒙 = 𝟑 𝟐 𝟒𝒙= 𝐥𝐧 𝟑 𝟐 𝒙= 𝟏 𝟒 𝐥𝐧 𝟑 𝟐 4=𝑙𝑛 3𝑥 𝒆 𝟒 =𝟑𝒙 𝒙= 𝟏 𝟑 𝒆 𝟒 E4 2 ln 𝑥 +1=5 𝟐 𝐥𝐧 𝒙 =𝟒 𝐥𝐧 𝒙 =𝟐 𝒙= 𝒆 𝟐 2𝑒 −𝑥 + 𝑒 𝑥 −3=0 𝟐+ 𝒆 𝒙 𝟐 −𝟑 𝒆 𝒙 =𝟎 𝒆 𝒙 𝟐 −𝟑 𝒆 𝒙 +𝟐=𝟎 𝒆 𝒙 −𝟏 𝒆 𝒙 −𝟐 =𝟎 𝒆 𝒙 =𝟏 𝒐𝒓 𝒆 𝒙 =𝟐 𝒙=𝟎 𝒐𝒓 𝒙=𝒍𝒏 𝟐 2 𝑥 𝑒 𝑥 =3 𝒍𝒏 𝟐 𝒙 𝒆 𝒙 =𝒍𝒏 𝟑 𝒍𝒏 𝟐 𝒙 +𝒍𝒏 𝒆 𝒙 =𝒍𝒏 𝟑 𝒙 𝒍𝒏 𝟐+𝒙=𝒍𝒏 𝟑 𝒙 𝒍𝒏 𝟐+𝟏 =𝒍𝒏 𝟑 𝒙= 𝒍𝒏 𝟑 𝒍𝒏 𝟐+𝟏 ? ? E2 ? E5 ? E6 ? E3 ?

Test Your Understanding Edexcel C3 June 2012 Q6 ? 𝑓 𝑥 >2 𝑓 𝑔 𝑥 =𝑓 ln 𝑥 =𝑒 ln 𝑥 +2=𝑥+2 𝑒 2𝑥+3 +2=6 2𝑥+3= ln 4 𝑥= −3+ ln 4 2 𝑦= 𝑒 𝑥 +2 𝑦−2= 𝑒 𝑥 𝑥= ln 𝑦−2 𝑓 −1 𝑥 = ln 𝑥−2 ? e) 𝑦=𝑓(𝑥) 3 ? ? 3 ? 𝑦= 𝑓 −1 (𝑥)

More Practice ? ? ? ? ? ? ? Find the exact solution to 3 𝑥 𝑒 2𝑥+1 =5 4 𝒍𝒏 𝟑 𝒙 𝒆 𝟐𝒙+𝟏 =𝒍𝒏 𝟓 𝒍𝒏 𝟑 𝒙 +𝒍𝒏 𝒆 𝟐𝒙+𝟏 =𝒍𝒏 𝟓 𝒙 𝐥𝐧 𝟑 +𝟐𝒙+𝟏= 𝐥𝐧 𝟓 𝒙= 𝐥𝐧 𝟓 −𝟏 𝟐+ 𝐥𝐧 𝟑 4 1 June 2007 Q1 ? 𝒙=𝟐 𝒙= 𝐥𝐧 𝟑 ? ? 2 Jan 2007 Q6 Jan 2007 Q6 𝒚=𝒍𝒏 𝟒−𝟐𝒙 𝒆 𝒚 =𝟒−𝟐𝒙 𝟐𝒙=𝟒− 𝒆 𝒚 𝒙=𝟐− 𝟏 𝟐 𝒆 𝒚 𝒇 −𝟏 𝒙 =𝟐− 𝟏 𝟐 𝒆 𝒙 ? 𝑓:𝑥→ ln 4−2𝑥 , 𝑥<2 𝑎𝑛𝑑 𝑥∈ℝ Domain: 𝒙∈ℝ ? 3 June 2006 Q4 ? 𝑇=425 300=400 𝑒 −0.05𝑡 +25 𝑒 −0.05𝑡 =0.6875 −0.05𝑡= ln 0.6875 𝑡=− 1 0.05 ln 0.06875 =7.49 ?

Exercises ? ? ? ? ? ? ? ? ? ? ? ? ? 5 Exercise 3B 1 f i 6 3 4 a ? b c 𝑓 𝑥 = ln 2+3𝑥 𝑥∈ℝ, 𝑥>𝑎 State the value of 𝑎. 𝟐+𝟑𝒙>𝟏 𝒙>− 𝟏 𝟑 Find the value of 𝑠 for which 𝑓 𝑠 =20 𝐥𝐧 𝟐+𝟑𝒔 =𝟐𝟎 𝟐+𝟑𝒔= 𝒆 𝟐𝟎 𝒔= 𝒆 𝟐𝟎 −𝟐 𝟑 ℎ 𝑥 =40−10 𝑒 3𝑥 𝑥>0, 𝑥∈ℝ State the range of the function. 𝒉 𝒙 <𝟒𝟎 Find the exact coordinate of the point where the graph intercepts the 𝑥-axis, in terms of ln 2 . 𝟐 𝟑 𝐥𝐧 𝟐 ,𝟎 Find ℎ −1 𝑥 stating its domain. 𝒉 −𝟏 𝒙 = 𝟏 𝟑 𝐥𝐧 𝟒− 𝟏 𝟏𝟎 𝒙 Domain is 𝑥<40 (i.e. range of ℎ) Solve ln 2 + ln 𝑥 =4 𝐥𝐧 𝟐𝒙 =𝟒 →𝒙= 𝟏 𝟐 𝒆 𝟒 Solve 𝑒 𝑥 + 𝑒 −𝑥 =2 𝒙=𝟎 Solve 3 𝑥 𝑒 2𝑥+1 =4 𝒙= 𝐥𝐧 𝟒 −𝟏 𝟐+ 𝐥𝐧 𝟑 Exercise 3B ? Solve (giving exact solutions) ln 2𝑥+1 =5 → 𝒙= 𝒆 𝟓 −𝟏 𝟐 2 𝑒 4𝑥 −3=8 →𝒙= 𝟏 𝟒 𝐥𝐧 𝟏𝟏 𝟐 Sketch the following: 𝑦= ln 𝑥+1 𝑦= ln 𝑥 2 𝑦= ln 4−𝑥 𝑦=3+ln⁡(𝑥+2) The price of a new car varies according to 𝑃=15 000 𝑒 − 𝑡 10 where P is price and t is age in years. State its value when new: £𝟏𝟓 𝟎𝟎𝟎 Calculate its age after 5 years. £9098 Find its age when the price falls below £5000 𝟓𝟎𝟎𝟎=𝟏𝟓 𝟎𝟎𝟎 𝒆 − 𝒕 𝟏𝟎 𝟏𝟏 𝒚𝒆𝒂𝒓𝒔 1 ? f ? ? i 6 3 ? ? 4 ? a ? ? b ? ? c ? ? ?