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Mathematics for Economists

Exponents and Logarithms Introduction

Many types of data can be modelled using exponential functions. Exponential functions naturally model the growth of a variable and are functions in which the variable 𝑥 appears as the exponent. For instance a financial variable that grows over time may have time 𝑡 as the exponent, that is 𝑓(𝑡) = 2𝑡. In this document we investigate the derivative of an exponential function. In finding the derivative of an exponential function we will discover that it describes the inverse of the exponential function which in turn is a logarithm. Understanding exponents and logarithms is crucial for working in applied economics as we often use logarithms to turn multiplicative relationships into additive relationships which are easier to manipulate. When working with data that exhibits exponential growth we often find that the graph is not a perfectly exponential due to volatility between data points. (This can occur for a number of reasons). However, many computer packages have curve fitting tools which allow us to easily develop equations that approximate data.

Exponential Functions

Properties of Exponents 1. if 𝑥 is positive integer then 𝑎𝑥 is 𝑎 multiplied by itself 𝑥 time 2. 𝑎𝑥 ∙ 𝑎𝑞 = 𝑎𝑥+𝑞 3. 𝑎𝑥 𝑎𝑞⁄ = 𝑎𝑥−𝑞 4. (𝑎𝑥)𝑞 = 𝑎𝑥𝑞 5. 𝑎0 = 1 6. 𝑎−𝑥 = 1 𝑎𝑥⁄ (if 𝑥 is a negative number then, 𝑎𝑥 is 1 𝑎|𝑥|⁄ , the reciprocal of 𝑎|𝑥|)

7. 𝑎1𝑛 = √an (the nth root of 𝑎)

8. 𝑎𝑚𝑛 = �√an �

m (the mth power of the nth root of 𝑎)

Exponential Growth Functions

The graph is the function 𝑓(𝑥) = 𝑦 = 2𝑥

x y-5 0.03125-4 0.0625-3 0.125-2 0.25-1 0.50 11 22 43 84 165 32

Exponential Functions

If 𝑎 is a real number with 𝑎 > 0 and 𝑎 ≠ 1, then the function 𝑓(𝑥) = 𝑎𝑥 is an exponential function with base 𝑎.

𝑦 = 2𝑥


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Notice that the 𝑥 axis is the horizontal asymptote for the graph. However unlike a rational function the graph approaches this asymptote in only a single direction. In the other direction the graph increases more rapidly than a polynomial. These graphs approach the 𝑥 axis but never actually ouch it. The steepness of the graph increases at an exponential rate. The larger the base the quicker the graph becomes asymptotic to the 𝑥 axis in one direction and steep in the other direction.

x a = 1.5 a = 2 a = 4 a = 6-5 0.131687 0.03125 0.000977 0.000129-4 0.197531 0.0625 0.003906 0.000772-3 0.296296 0.125 0.015625 0.00463-2 0.444444 0.25 0.0625 0.027778-1 0.666667 0.5 0.25 0.1666670 1 1 1 11 1.5 2 4 62 2.25 4 16 363 3.375 8 64 2164 5.0625 16 256 12965 7.59375 32 1024 7776

𝑦 = 1.5𝑥

𝑦 = 2𝑥 𝑦 = 4𝑥 𝑦 = 6𝑥

𝑦 = 𝑓(𝑥) = 𝐶(𝑎𝑥) (𝑎 > 1) Exponential Growth Functions

The graph represents the general form of exponential growth functions. Note that in general terms the 𝑦 intercept is 𝐶. If we just have 𝑎𝑥 , where 𝐶 = 1, the 𝑦 intercept is 1. Likewise if 𝐶 = 2 then the 𝑦 intercept will become 2.

(0,𝐶)


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Example 1: Exponential Growth Functions

If $5000 is invested at 8%, compounded monthly, then what is the future value of the investment after 5 years and 10 years ? Graph the function.

Firstly we divide 8% by 𝑚 = 12 months for the rate. The apply the following formula, where 𝑃 is the principle amount in year 1, 𝑆 is the future value in year 𝑛 and 𝑟 is the interest rate. So 𝑚𝑛 is the total number of compounding periods for which interest is applied. In this case there will be 60 periods for 5 years and 120 periods for 10 years.

𝑆 = 𝑃(𝑟 𝑚⁄ )𝑚𝑛

𝑆5 𝑦𝑒𝑎𝑟𝑠 = 5000(1.00667)12(5) = $7,449

𝑆10 𝑦𝑒𝑎𝑟𝑠 = 5000(1.00667)12(10) = $11,098

It is obvious from the graph and formula that compounding interest is essentially a exponential growth function.

Exponential Decay Functions

The graph is the function 𝑓(𝑥) = 𝑦 = �12�𝑥

= 2−𝑥

Exponentials whose bases are between zero and one have graphs different to those of exponential growth functions. If a function is of the form 𝑦 = 𝑏𝑥 where 0 < 𝑏 < 1 then using the properties of exponents we have

𝑦 = 𝑎𝑥 = �1𝑏�𝑥

= (𝑏𝑥)−1 = 𝑏−𝑥 = 𝑎−𝑥

x a = 0.5-5 32-4 16-3 8-2 4-1 20 11 0.52 0.253 0.1254 0.06255 0.03125

𝑦 = �12�𝑥

𝑆 = 𝑃(𝑟 𝑚⁄ )𝑚𝑛


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We have written 𝑦 = 𝑏𝑥 in the form 𝑦 = 𝑎−𝑥, where 𝑎 > 1 and 𝑦 = 𝑏𝑥. Therefore if we have 𝑦 = (1 2⁄ )𝑥, then

𝑦 = �12�𝑥

= (2𝑥)−1 = 2−𝑥

From the diagrams you can probably notice that the graphs are reflections. For instance the

graph of 𝑦 = �12�𝑥 is the reflection of 𝑦 = 2𝑥 in the y axis. Likewise 𝑦 = �1

3�𝑥 is the reflection

of 𝑦 = 3𝑥 and so forth.

x a = 1.5 a = 2 a = 3 a = 6-5 7.59375 32 243 7776-4 5.0625 16 81 1296-3 3.375 8 27 216-2 2.25 4 9 36-1 1.5 2 3 60 1 1 1 11 0.666667 0.5 0.333333 0.1666672 0.444444 0.25 0.111111 0.0277783 0.296296 0.125 0.037037 0.004634 0.197531 0.0625 0.012346 0.0007725 0.131687 0.03125 0.004115 0.000129

𝑦 = 𝑓(𝑥) = 𝐶(𝑎−𝑥) (𝑎 > 1)

Exponential Decay Functions

The graph represents the general form of exponential decay functions. Again, note that in general terms the 𝑦 intercept is 𝐶. If we just have 𝑎𝑥 , where 𝐶 = 1, the 𝑦 intercept is 1. Likewise if 𝐶 = 2 then the 𝑦 intercept will become 2.

𝑦 = 1.5𝑥

𝑦 = 2𝑥 𝑦 = 4𝑥 𝑦 = 6𝑥

(0,𝐶)


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A Note on Negative Bases

Negative bases are not valid for exponential functions. A negative integer for 𝑎 would take positive values for when 𝑥 is even and negative values for when 𝑥 is odd. However the function never takes on the value zero in between these values. In addition the square root of a negative number is not even defined. Essentially we only work with exponential functions.

Example 2: Evaluating Exponents

Evaluate the exponential function −2𝑥 at 𝑥 = 2 and 𝑥 = 3

−2(2) = 4

−2(3) = −8

By using points between 𝑥 = 2 to 𝑥 = 3 attempt to evaluate every 10 decimal increment between 2 and 3. Explain what you notice. (Hint: it is best to do this in excel)

As discussed above between even and odd numbers a function with a negative base will not be invalid and is not defined. So for 𝑥 = 2 we have 4 a positive number and for 𝑥 = 3 we have a negative number −8, yet it is never zero in between. This is why we are not allowed to have negative bases for the exponential function.

Example 3: Exponents

State whether each of the following statements is either true or false.

(a) 𝑥 can give a negative value for 𝑦 if 𝑦 = 𝑎𝑥 and 𝑎 > 1? (b)The graph of 𝑦 = 𝑎−𝑥, with 𝑎 > 1, approaches the positive 𝑥 axis as an asymptote.

(c) The graph of 𝑦 = �13�𝑥 is the graph of 𝑦 = 3𝑥 reflected about the 𝑦 axis.

(d) The 𝑦 intercept of the function 𝑦 = 2(3𝑥) is given at the point (0,2).

Solution

(a) False (the 𝑥 axis is the horizontal asymptote for the graph and therefore 𝑦 > 0) (b) True (the asymptote is the 𝑥 axis in the positive half) (c) True (the graph of these two functions are symmetrical) (d) True �𝑦 = 𝑓(𝑥) = 𝐶(𝑎𝑥)�

x a y2 -2 4

2.1 -2 #NUM!2.2 -2 #NUM!2.3 -2 #NUM!2.4 -2 #NUM!2.5 -2 #NUM!2.6 -2 #NUM!2.7 -2 #NUM!2.8 -2 #NUM!2.9 -2 #NUM!3 -2 -8


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The Number 𝒆

In this section we shall discuss the irrational number 𝑒. The exponential functions with base 𝑒 often arise in natural ways. To understand the irrational number 𝑒 we will examine the growth of a savings account from making an initial investment. Suppose that we deposit $𝐴 dollars into a savings account that pays a simple annual interest rate of 𝑟. This amount will grow to 𝐴 + 𝑟𝐴 = 𝐴(1 + 𝑟) in year 1. After two years this amount will be

𝐴(1 + 𝑟)(1 + 𝑟) = 𝐴(1 + 𝑟)2

dollar in the account. Likewise After 𝑡 years the there will be

𝐴(1 + 𝑟)𝑡

dollars in the account. Now think of the situation when we are compounding more frequently, such as semi annual, quarterly and monthly. For quarterly we are compounding four times per year so we the interest applied in each quarter is 𝑟/4. After the first quarter the account will have 𝐴 + 𝐴 �𝑟

4� = 𝐴(1 + 𝑟

4). After the second quarter we will have 𝐴(1 + 𝑟

4)2

and after the fourth quarter 𝐴(1 + 𝑟4)4. Therefore the amount of dollars in the account after t

years we be 𝐴(1 + 𝑟4)4𝑡.

Generally, if interest is compounded 𝑛 times per year there will be 𝐴(1 + 𝑟𝑛

)1 after one period

of compounding, 𝐴(1 + 𝑟𝑛

)𝑛𝑡 after 𝑡 years of compounding.

Continuously Compounding

We know that banks and financial institutions offer rates with various compounding such as fortnightly, daily and continuously. If interest is compounded so frequently we are concerned with the limit of (1 + 𝑟

𝑛)𝑛 as 𝑛 approaches infinity. For the purpose of intuition we can

calculate an interest rate of 100% for different frequencies of compounding.

The tables shows an increasing sequence in 𝑛 which converges to approximately 2.718. It turns out that this number is irrational and cannot be written as a fraction or a repeating decimal. We use the letter 𝑒 to denote this number. In addition this number plays an important role in economics and finance. We call the following function the exponential function.

𝑓(𝑥) = 𝑒𝑥


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Graph of Exponential Function


𝑓(𝑥) = 𝑒−𝑥

Example 4: Exponential Functions

Suppose a financial institution offers continuously compounded interest rates on deposits of $100 or more. How much will $100 be at 5% per annum continuously compounded for 3, 5 and 10 years?

x e^x-7 0.001-6 0.002-5 0.007-4 0.018-3 0.050-2 0.135-1 0.3680 11 32 73 204 555 1486 4037 1097

x e^x-7 1096.6-6 403.4-5 148.4-4 54.6-3 20.1-2 7.4-1 2.70 1.01 0.3682 0.1353 0.0504 0.0185 0.0076 0.0027 0.001


lim𝑛→∞

�1 +1𝑛�𝑛

= 𝑒𝑟

lim𝑛→∞

�1 +1𝑛�𝑛𝑡

= lim𝑛→∞

��1 +1𝑛�𝑛

�𝑡

= (𝑒𝑟)𝑡 = 𝑒𝑟𝑡

The Exponential Function

If we let the account grow for 𝑡 years then

If we deposit 𝐴 dollars into an account which pays an annual interest rate of 𝑟 compounded continuously, then after 𝑡 years the account grows to 𝐴𝑒𝑟𝑡 dollars.


𝑓(𝑥) = 𝑒−𝑥


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To solve this we use the formula 𝐴𝑒𝑟𝑡

If we let the account grow for 3, 5 and 10 years then

𝐴𝑒𝑟𝑡 = 100𝑒0.05(3) = 116.18

𝐴𝑒𝑟𝑡 = 100𝑒0.05(5) = 128.40

𝐴𝑒𝑟𝑡 = 100𝑒0.05(10) = 164.87

If we deposit $100 dollars into an account which pays an annual interest rate of 5% compounded continuously, then after 3 years the account grows to $116.18 dollars, after 5 years the account grows to $128.40 and after 10 year the account will be $164.87.

Example 5: Exponential Functions

Suppose a portion of the population, such as pensioners, are on a fixed income. If the average fixed income for a pensioner is fixed $25000 per year then a natural question to ask is what is the purchasing power of this amount. Each year the amount of goods and services and that can be purchased will decrease. Suppose purchasing power will decrease at the rate of inflation of 3% and purchasing power is modelled by the function

𝑃𝑃 = 25000𝑒−0.03𝑡

Then find the purchasing power of $25000 in 3 and 10 years and graph the function.

𝑃𝑃 = 25000𝑒−0.03(3) = 22848

𝑃𝑃 = 25000𝑒−0.03(10) = 18520

Therefore in 3 years the $25000 will only be able to purchase $22848 worth of goods. In 10 years time the purchasing power of $25000 today will be $18250. This is an example of how inflation will erode fixed income assets. The function is graphed below and is a exponential decay function.

Logarithms

A general exponential function 𝑦 = 𝑎𝑥, with a base 𝑎 > 1 will be strictly increasing. As the function is strictly increasing it will have a natural inverse. The inverse of a function is found by solving 𝑦 = 𝑓(𝑥) for 𝑥 in terms of 𝑦. For instance the inverse of an increasing linear function 𝑓(𝑥) = 𝑎𝑥 + 𝑏 is the linear function 𝑔(𝑦) = �1

𝑎� (𝑦 − 𝑏), which is computed by

solving the linear equation 𝑦 = 𝑎𝑥 + 𝑏 for 𝑥 in terms of 𝑦

𝑓(𝑥) = 25000𝑒−0.03(𝑡)


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𝑦 = 𝑎𝑥 + 𝑏 ⟹ 𝑥 = 1𝑎

(𝑦 − 𝑏)

Essentially the inverse function 𝑔 of 𝑓 reverses the operation of 𝑓 such that

𝑔�𝑓(𝑥)� = 𝑥

(Inverse functions are covered in Introduction to Functions)

We are unable to compute the inverse of the increasing exponential function 𝑓(𝑥) = 𝑎𝑥 explicitly because we cannot solve 𝑦 = 𝑎𝑥 for 𝑥 in terms of 𝑦 as we did with the example above. Instead we call the inverse of this function the base 𝑎 logarithm and denote it as

𝑦 = 𝑙𝑜𝑔𝑎(𝑧) ⟹ 𝑎𝑦 = 𝑧

The logarithm of 𝑧, is the power to which one must raise 𝑎 to yeild 𝑧 and therefore

𝑎𝑙𝑜𝑔𝑎(𝑧) = (𝑧) and 𝑙𝑜𝑔𝑎(𝑎𝑧) = 𝑧

Example 6: Logarithmic Functions

Write the following in logarithmic form.

(a) 125 = 53

(b) 64 = 43

(c) 9 = 32

(d) 127

= 3−3

Solution

(a) 3 = log5 125

(b) 3 = log4 64

(c) 2 = log3 9

(d) −3 = log3( 127

)

𝑦 = 𝑙𝑜𝑔𝑎𝑥

𝑎𝑦 = 𝑥

Logarithmic Functions

If 𝑎 > 0 and 𝑎 ≠ 1, the logarithmic function

has domain 𝑥 > 0 and base 𝑎 and is defined by

in exponential form. The 𝑎 is called the base in both 𝑦 = 𝑙𝑜𝑔𝑎𝑥 and 𝑎𝑦 = 𝑥. However 𝑦 is the logarithm in 𝑦 = 𝑙𝑜𝑔𝑎𝑥 and the exponent in 𝑎𝑦 = 𝑥. Therefore the logarithm is an exponent.


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Write the following in exponential form.

(a) log5 125 = 3

(b) log7 343 = 3

(c) log6 �136� = −2

Solution

(a) 125 = 53

(b) 343 = 73

(c) 136

= 6−2

Logarithms of Base 10

The logarithmic function with base 𝑎 = 10 is very commonly used and is denoted as 𝑦 = 𝐿𝑜𝑔𝑥.

𝑦 = 𝐿𝑜𝑔𝑥 ⟹ 10𝑦 = 𝑧

The Log of 100 is the power of 10 which yields 100. So as 102 = 100, 𝐿𝑜𝑔1000 = 3. Likewise the Log of 100,000 is given by 𝐿𝑜𝑔100,000 = 5 because 105 = 100,000. Also the Log of 845 is given by 𝐿𝑜𝑔845 = 2.93 because 102.93 = 845.

Log1 = 0 100 = 1 Log10 = 1 101 = 10 Log1000 = 3 103 = 1000 Log100000 = 5 105 = 100000 Log29 = 1.46 101.46 = 29 Log845 = 2.93 102.93 = 845

When we graph 𝑦 = 𝐿𝑜𝑔𝑥 and 𝑦 = 10𝑥 we discover that they are the reflection of each other. This is because they are inverse functions. The negative 𝑥 axis will be the horizontal asymptote for 𝑦 = 10𝑥 and the negative 𝑦 axis is a vertical asymptote for the 𝑦 = 𝐿𝑜𝑔𝑥.

Logarithms of Base 𝒆

The exponential function 𝑓(𝑥) = 𝑒𝑥 has an inverse which is known as the natural logarithm. Formally we write this as

ln𝑥 = 𝑦 ⟹ 𝑒𝑦 = 𝑥

ln𝑥 is the power to which one must raise 𝑒 to get 𝑥. In general terms we can summarise the definition as the following.

𝑒ln𝑥 = 𝑥 and ln 𝑒𝑥 = 𝑥


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When we graph 𝑦 = ln𝑥 and 𝑦 = 𝑒𝑥 we discover that they are the reflection of each other. This is because they are inverse functions. The negative 𝑥 axis will be the horizontal asymptote for 𝑦 = 𝑒𝑥 and the negative 𝑦 axis is a vertical asymptote for the 𝑦 = ln𝑥.

ln 𝑒 = 1 𝑒1 = 𝑒 ln 0.1 = -2.3 𝑒−2.3 = 0.1 ln 1 = 0 𝑒0 = 1 ln 2 = 0.6931 𝑒0.6931 = 2

ln 20 = 2.9957 𝑒2.99 = 20


Compute the following

(a) 𝐿𝑜𝑔200, (b) 𝐿𝑜𝑔 𝑒, (c) ln 15, (d) 𝐿𝑜𝑔10, (e)ln 2.87

Solution (a) 2.3 (b) 0.434 (c) 2.708 (d) 1 (e) 1.022


For what value of 𝑥 is 𝑦 = 𝑙𝑜𝑔𝑎𝑥 , 𝑎 > 0, 𝑎 ≠ 1, defined?

Solution

Notes on the graphs of logarithmic functions

𝑦 = 𝑙𝑜𝑔𝑎𝑥 , 𝑎 > 0

1. The 𝑥 intercept will be at the point (1,0)

2. The domain of the is the set of all positive real numbers

3. The range of the function is the set of all real numbers

4. The Asymptote occurs on the negative half of the 𝑦 axis


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For values of 𝑥 > 0 the value of 𝑦 = 𝑙𝑜𝑔𝑎𝑥 is defined. To see this we can graphing the function 𝑦 = ln𝑥 as follows.

It is important to note that the domain is positive for all real values of 𝑥 and the asymptote is on the negative half of the 𝑦 axis.


Consider a firm which is producing a new smart phone into the market. Suppose the marketing team have run some simulations and estimate that in 𝑚 months prior to its market share being 𝑠 % the products share of the smart phone market. Find when the product will have a 50% share of the market.

𝑚 = 15 ln �60

60 − 𝑠�

Solution

The market share of 50% means that we set 𝑠 = 50 and calculate the right hand side of the equation.

𝑚 = 15 ln �60

60 − 50� = 15 ln �

6010� = 15 ln 6 = 26.87

Therefore it will take 26.87 months for the product to have a 50% share of the smart phone market.

Properties of Logarithms

The five properties below are essentially the same as the properties of exponents that we presented previously. We prove these properties on the right hand side of the properties by letting 𝑢 = log𝑎 𝑟 and 𝑣 = log𝑎 𝑠, and therefore letting 𝑟 = 𝑎𝑢 and 𝑠 = 𝑎𝑣.

1. log(𝑟 ∙ 𝑠) = log 𝑟 + log 𝑠 = log(𝑎𝑢 ∙ 𝑎𝑣) = log(𝑎𝑢+𝑣) = 𝑢 = 𝑣 = log 𝑟 + log 𝑠

2. log(1 𝑠⁄ ) = − log 𝑠 = log � 1𝑎𝑣� = log(𝑎−𝑣) = −𝑣 = − log 𝑠

3. log(𝑟 𝑠⁄ ) = log 𝑟 − log 𝑠 = log(𝑎𝑢 𝑎𝑣⁄ ) = log(𝑎𝑢−𝑣) = 𝑢 − 𝑣 = log 𝑟 − log 𝑠

4. log 𝑟𝑠 = 𝑠 log 𝑟 = log(𝑎𝑢)𝑠 = log𝑎𝑢𝑠 = 𝑢𝑠 = 𝑠 log 𝑟

5. 𝑙𝑜𝑔1 = 0 see paragraph below


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The exponential form of 𝑦 = log𝑎 𝑥 is 𝑎𝑦 = 𝑎𝑥, so 𝑦 = 𝑥, and therefore log𝑎 𝑥 = 𝑥

Note that the fifth property of logs follows directly from the fifth property of 𝑎𝑥 and the fact that 𝑎𝑥 and 𝑙𝑜𝑔𝑎 are inverses of each other.

Example 11: Properties of Logarithms

Solutions

(a) 2log2 4 = 4

(b) ln 𝑒𝑥 = log𝑒 𝑒𝑥 = 𝑥

(c) 𝑒ln𝑥 = 𝑥

(d) log2 �416� = log2 4 − log2 16 = 2 − 4 = −2

(e) ln �79� = ln 7 − ln 9 = −0.2513

(f) 2𝑥−2 = 5 → ln 2𝑥−2 = ln 5 → (𝑥 − 2) ln 2 = ln 5 ∴ 𝑥 = 2 + �ln5ln2�

(g) 2𝑥 = 𝑒5 → 𝑥 ln 2 = ln 𝑒5 → 𝑥 ln 2 = 5 ∴ 𝑥 = � 5ln2�

(h) ln 77 = ln(7 ∙ 11) = ln 7 + ln 11 = 4.3438

(i) log3(9 ∙ 27) = log39 + log327 = 2 + 3 = 5 (check: log3(243) = 5 because 35 = 243)

(j) ln√𝑒 = 0.5


How long does it take for $3000 to grow to $5000 if the interest rate is 4% per annum is compounded continuously?

𝑆 = 𝐴𝑒𝑟𝑡

5000 = 3000𝑒(0.04)(𝑡)

ln �50003000

� = 𝑙𝑛�𝑒(0.04)(𝑡)�

0.5108 = 0.04𝑡

𝑡 = 12.77

(a) Simplify 2log2 4 (f) Solve 2𝑥−2 = 5 for 𝑥

(b) Simplify ln 𝑒𝑥 (g) Solve 2𝑥 = 𝑒5 for 𝑥

(c) Simplify 𝑒ln𝑥 (h) Find ln 77

(d) Evaluate log2 �416� (i) Find log3(9 ∙ 27)

(e) Evaluate ln �79� (j) Evaluate ln√𝑒


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Therefore it will take 12.77 years for $3000 to become $5000 at 4% continuously compounded.

Derivatives of Exponents and Logarithms

As logarithmic and exponential functions are frequently used in economic analysis we must also know the derivatives of these functions. For the natural logarithm and exponential function the derivatives are quite simple. The functions 𝑒𝑥 and ln𝑥 are both continuous functions on their domains and for derivatives of every order. From the graphs of these functions we can see that they are continuous functions.

For the exponential function the first derivative is given by

(𝑒𝑥)′ = 𝑒𝑥

also if 𝑢(𝑥) is a differentiable function,

�𝑒𝑢(𝑥)�′

= �𝑒𝑢(𝑥)� ∙ 𝑢′𝑥

For the logarithmic function the first derivative is given by

(ln𝑥)′ =1𝑥

and if 𝑢(𝑥) is a differentiable function,

((ln𝑢(𝑥))′ =𝑢′(𝑥)𝑢(𝑥)

𝑖𝑓 𝑢(𝑥) > 0

See the appendix for the proofs of these properties.

Example 13: Computing Derivatives for Logarithmic and Exponential Functions

Compute the following derivatives

(a) 𝑒9𝑥 (b) 𝑒3𝑥3 (c) 𝑒𝑥 ln𝑥 (d)ln𝑥5 (e) ln(𝑥3 + 𝑥 + 3) (f) 𝑥𝑒2−𝑥

Solution

(a) (𝑒9𝑥)′ = 9𝑒9𝑥

(b) �𝑒3𝑥3�′

= 9𝑥2 𝑒3𝑥3

(c) (𝑒𝑥 ln𝑥) = 𝑒𝑥 ln 𝑥 + 𝑒𝑥

𝑥

(d) (ln𝑥5)′ = 1𝑥5∙ 5𝑥4 = 5

𝑥

(e) (ln(𝑥3 + 𝑥 + 3))′ = 2𝑥+3𝑥3+𝑥+3

(f) (𝑥𝑒2−𝑥)′ = 𝑒2−𝑥 − 𝑥𝑒2−𝑥

The Derivative of 𝒚 = 𝐥𝐨𝐠𝒂 𝒙: If we have a logarithmic function with a base other than 𝑒 then we use the change of base formula to express it as a natural logarithm.

(a)

(b)

(c)

(d)


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log𝑎 𝑥 =ln𝑥ln𝑎

This means that derivatives of functions with bases other than 𝑒 will have the constant 1/ ln𝑎 as a factor. This is best shown in the example below, where 1/ ln 3 is the constant factor.

Example 14: Derivatives of Logarithmic Functions

Find the derivative of the following function.

𝑦 = log3(𝑥5 + 2)

𝑦 = log3(𝑥5 + 2) =ln(𝑥5 + 2)

ln 3=

1ln 3

∙ ln(𝑥5 + 2)

Thus,

𝑑𝑦𝑑𝑥

=1

ln 3∙

1𝑥5 + 2

∙ 3𝑥2 =3𝑥2

ln 3(𝑥5 + 2)



𝑦 = ln[𝑥(𝑥7 − 2)8]

𝑦 = ln𝑥 + ln(𝑥7 − 2)8

𝑦 = ln 𝑥 + 8 ln(𝑥7 − 2)

𝑑𝑦𝑑𝑥

=1𝑥

+56𝑥6

𝑥7 − 2



𝑦 = log5(𝑥2 + 3)

𝑦 = log5(𝑥2 + 3) =ln(𝑥2 + 3)

ln 5=

1ln 5

∙ ln(𝑥2 + 3)

𝑑𝑦𝑑𝑥

=1

ln 5∙

1𝑥2 + 3

∙ 2𝑥 =2𝑥

(𝑥2 + 3) ln 5

Some Applications of Exponents and Logs

In economics and finance we are often concerned with knowing the quantum, or value of something. Logarithmic and exponential functions provide us with the means to work out how much something is worth a point in time. In the examples above we have shown that the future value of an amount of money that is continuously compounded can be found easily using the exponential functions. Also by using logarithmic functions we are able to find out how long it takes for a sum of money to grow to a certain amount. Now we know


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how to differentiate exponential and logarithmic functions we can find optimal values for these functions. This is very useful in financial and economic analysis.

Present Values: If 𝐴 is invested at 𝑟 for 𝑡 years then the future value 𝑆 is given by the formula 𝑆 = 𝐴𝑒𝑟𝑡. Alternatively to generate 𝑆 dollars in 𝑡 years from today at 𝑟 we need to invest 𝐴 = 𝑆𝑒−𝑟𝑡. In other words the present value of 𝑆 is given by,

𝐴 = 𝑆𝑒−𝑟𝑡 Optimal Investment Horizon: The optimal horizon of an investment refers to the optimal holding time of an asset. There are many assets which appreciate and depreciate over time such as vintage wine, real estate, forestry plantations and mining to name a few. In evaluating the optimal present value of an investment depends on the alternative investments and we usually compare the investment to the interest rate. For instance if a person was to invest in real estate then the growth in the value of the real estate investment would be compared to growth rate of money in the bank. The investor will switch investments at the point where the growth rates cross paths.

Consider Peter who owns real estate that is expected to have a market value of 𝑉(𝑡) in 𝑡 years from now. The interest rate is expected to be 8% in the foreseeable future. Suppose the function for this investment is given by

𝑉(𝑡) = 20,000𝑒√𝑡

The optimal investment horizon is given by maximising the present value

𝐹(𝑡) = 20,000𝑒√𝑡𝑒−0.08𝑡 = 20,000𝑒√𝑡−0.08𝑡

To find the optimal investment horizon we take the first derivative set to zero and solve for 𝑡.

0 = 𝐹′(𝑡) = 20,000𝑒√𝑡−0.08𝑡 �1

2√𝑡− 0.08�

12√𝑡

= 0.08 𝑜𝑟 𝑡 = �1

0.16�2

= 39.065

Therefore the optimal time to sell the real estate is 39.065 years.


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(ln𝑥)′ =1𝑥

ln(𝑥 + ℎ) − ln𝑥ℎ

=1ℎ

ln �𝑥 + ℎ𝑥

� = ln �1 +ℎ𝑥�1/ℎ

= ln �1 +1/𝑥1/ℎ

�1/ℎ

limℎ→0

ln(𝑥 + ℎ) − ln𝑥ℎ

= lim𝑚→∞

ln�1 +1𝑥𝑚�

𝑚

= ln 𝑒1𝑥 =

1𝑥

((ln𝑢(𝑥))′ =𝑢′(𝑥)𝑢(𝑥)

𝑖𝑓 𝑢(𝑥) > 0

𝑑𝑑𝑥

(ln𝑢(𝑥)) =𝑢′(𝑥)𝑢(𝑥)

((ln𝑢(𝑥))′ =1

𝑢(𝑥)∙ 𝑢′(𝑥) =

𝑢′(𝑥)𝑢(𝑥)

(𝑒𝑥)′ = 𝑒𝑥

Appendix

Proofs: Derivatives of Exponential and Logarithmic Functions

The most intuitive way of proving (a), (b), (c) and (d), below, is to start with (c) then (d) then (a) and (b).

(a) (𝑒𝑥)′ = 𝑒𝑥

(b) �𝑒𝑢(𝑥)�′

= �𝑒𝑢(𝑥)� ∙ 𝑢′𝑥

(c) (ln𝑥)′ = 1𝑥

(d) ((ln𝑢(𝑥))′ = 𝑢′(𝑥)𝑢(𝑥)

𝑖𝑓 𝑢(𝑥) > 0

Proof for (c)

Now let 𝑚 = 1/ℎ. As ℎ → 0,𝑚 → ∞, then

Therefore (ln 𝑥)′ = 1𝑥. Note that because we can interchange the limit in the above

expressions if follows that 𝑦 = ln𝑥 is a continuous function.

Proof for (d)

To prove this theorem we use the chain rule. If 𝑢(𝑥) is a differentiable function then

Now simply apply the chain rule to the composite function 𝑓(𝑥) = ln𝑢(𝑥). The derivative of the outside function ln𝑢(𝑥) is 1/𝑢(𝑥) which is multiplied by 𝑢′(𝑥) which is the derivative of the inside function 𝑢.

Proof for (a)

To prove this lemma we simply to take the derivatives of both sides of the equation and apply the chain rule. First we use the definition of ln𝑥 so we can write ln 𝑒𝑥.


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(ln 𝑒𝑥)′ =

1𝑒𝑥

∙ (𝑒𝑥)′ = 1

�𝑒𝑢(𝑥)�′

= �𝑒𝑢(𝑥)� ∙ 𝑢′𝑥

�𝑒𝑢(𝑥)�′

= �𝑒𝑢(𝑥)� ∙ 𝑢′𝑥

Appendix

Proof for (b)

Again we use the chain rule to prove this statement. We define the function as 𝑦 = 𝑒𝑢(𝑥), where the outside function is 𝑒𝑘 and the inside function is 𝑒𝑢(𝑥). The derivative of the outside function is 𝑒𝑘′ = 𝑒𝑘 and the derivative of the inside function is 𝑢′𝑥. Therefore we conclude that

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