# Exponential and Logarithmic Functions. Professor Peter Cramton Economics 300

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1 Exponential and Logarithmic Functions Professor Peter Cramton Economics 300

2 Exponential functions Modeling growth Constant percentage growth per unit time Logarithmic functions Inverse of exponential functions

3 Interest rate r Growth of money Value of X t after 1 time period: X t+1 = (1 + r)x t r = 10%; \$10 today is worth (1.1)10 = \$11 next year Value of X t after 2 time periods: X t+2 = (1 + r)x t+1 = (1 + r)(1 + r)x t = (1 + r) 2 X t Value of X t after n time periods X t+n = (1 + r)x t+n-1 = (1 + r) n X t \$1 earning 5% for 50 years = \$11.47 \$1 earning 10% for 50 years = \$ Doubling interest rate has a huge impact

4 Exponential growth 50 (1 r) n r=10% r=5% n (years)

5 More frequent compounding Once per year: Twice per year: k times per year: times per year: (1 r) r 2 r k (1 ) 2 k 1 r r 1 k r m r m r (1 ) (1 ) (1 ) 1 m lim(1 ) m (1 1) k k k k e mr

6 Most important constant in economics e 1) k k lim(1 k

7 50 40 Exponential growth Continuous compounding r=10% rn e Annual compounding (1 r) n r=5% n (years)

8 Effective rate vs. annual rate Annual rate of r A = 10% with continuous compounding What is effective rate r E over the year? 1 r E 1 r A = 10% then r E = 10.52% (continuous compounding) r A = 10% then r E = 10.50% (daily compounding) r E e r A e r A 365 r A r (1 ) 1 E

9 The Mating Game A surprising application of e [See mating-game.nb]

10 Present value (discrete) What is \$1 next year worth today With r = 10%, than \$1 today is worth \$1.10 next year X X t1 t X (1 r) X 1 t1 r X X tn X (1 r) t t n (1 r) n t n

11 Present value (continuous) What is \$1 next year worth today With r = 10%, than \$1 today is worth \$1.10 next year X t1 X r e X X t1 X e r t r t1 e X X tn X e rn t rn tn e t

12 Net present value Investments generate costs and revenues over time What is the value today of the sequence of cash flows from an investment? CF NPV Revenue t0 Cost t t t NPV CF CF CF CF 1... n t 0 1 r (1 r) n (1 r) n t CF t t0 1, where = 1 1 r n t

13 Examples Discount rate r 0 1 Discount factor 1 1 r Perpetuity: value of \$1 each period forever 1 d 1 Annuity: value of \$1 each period for n periods d n 1 1 n

14 Example Discount rate r 0 1 Discount factor 1 1 r Perpetuity: value of \$1 each period forever d d 2 (1 ) d 1 d , subtracting yields r = 10%; =.909; perpetuity = \$11.00

15 Example Discount rate r 0 1 Discount factor 1 1 r Annuity: value of \$1 each period for n periods d n n1 2 n d..., subtracting yields n (1 ) d 1 d n n n 1 1 n r = 10%; =.909; 20-year annuity = \$9.36

16 Logarithms Inverse of exponential function y log ( x) finds exponent y such that b x b b is the base Most commonly b = 10 or b = e log base e is called natural logarithm: y y y ln( x) log( x)

17 Log is inverse of exponential function

18 Log base 10 Example of log base 10 x y x y 10 y log ( x) 10 Examples of log scales Shock waves (Richter scale for earthquakes) (2011: Virginia 5.8, Japan 9.0; 1585 times larger) Sound waves (decibels for sound) Radio waves (Hz, khz, MHz, GHz)

19 Log base 10 and base 2

20 10 5 Log plot of exponential growth y 10 x y e x y 2 x

21 Properties of logarithmic functions log b 1 log xy log x log y x log log xlog y log log b b b b b b b b b y x y log x x log log a a x b b y

22 Natural logs (base e) Continuous growth models Same properties hold ln e 1 ln xy ln x ln y x ln ln xln y y ln y x y ln x Example: Yahoo Finance (plotting stock history)

23 DJ vs. S&P vs. Nasdaq (linear scale)

24 DJ vs. S&P vs. Nasdaq (log scale)

25 Average return from stocks return r Dow Jones \$ in February 1978 Dow Jones \$ in February r e ln(742.12) 30r ln( ) ln( ) ln(742.12) r %

26 Average return from stocks return r Dow Jones \$ in February 1978 Dow Jones \$ in February r e ln(742.12) 32r ln( ) ln( ) ln(742.12) r 8.10% 32

27 Average return from stocks return r Dow Jones \$ in February 1978 Dow Jones \$15, in February r e ln(742.12) 36r ln( ) ln( ) ln(742.12) r 8.50% 36

28 Average return from stocks return r, accounting for inflation Dow Jones \$ in Feb 1978; CPI 62.5 Dow Jones \$15, in Feb 2014; CPI r ( / 62.5) e / ln( / 62.5) 36r ln( / 234.1) ln( / 234.1) ln( / 62.5) r %

29 Average growth rate Value at time 0: V 0 Value at time T: V T Assume constant percentage growth per unit time V e 0 rt ln( V ) rt ln( V ) r 0 V T ln( V ) ln( V ) T T 0 T

30 Inflation in Mexico,

31 Cobb-Douglas production One special case: In general: Q y AL K x x 1 2 Q = real GDP L = labor K = capital are parameters and

32 Cobb-Douglas production We measure Q, L, and K at each time: Taking logs: Q A L K t t t t ln Q ln A ln L ln K t t t t Nice linear model! Can estimate parameters with econometrics Using subtraction lnq ln Q (ln A ln A ) (ln L ln L ) (ln K ln K ) t t1 t t1 t t1 t t1 What is? ln Q t ln Qt 1

33 How long does it take for something to V e 0 rn rn double? e V / V 2 rn n V n With r = 10% it takes 7 years for value to double With r = 5% it takes 14 years for value to double Moore s Law of electronics: a doubling every 18 months r =.6931/1.5 = 46% 0 ln( e ) ln(2) n ln(2).6931 r r

34

35 Properties of logarithmic functions log b 1 log xy log x log y x log log xlog y log log b b b b b b b b b y x y log x x log log a a x b b y

36 Simplify 10 ln log ln log (100) e x e 5 1 x e ln x x y 100 xx log 2 0 x 5 5log x 5 2( ln x ln y)

37 Problem \$10,000 invested at 5% with continuous compounding When do you have \$15,000? Use formula for future value: X t+n = X t e rn 15,000 = 10,000 e.05n Solve for n e.05n = 15,000/10,000 = n = ln(1.5) n = ln(1.5)/.05 =

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