Summary 2
Exponential and logarithmic functions
Test 1: 13/03/24
Sit in – 240/011
Time: 1800-2000 hours
Introduction
 General properties of:
 𝑦 = 𝑓 𝑥 = 𝑎 𝑥 ; 𝑜𝑟 𝑡; 𝑦 = 𝑓 𝑡 = 𝑎𝑡 , 𝑎 > 0 𝑎𝑛𝑑 𝑎 ≠ 1
 Domain of the function – is the set of all real numbers, i.e.,
the variable t can take negative numbers
 Range of the function – is the set of all positive real
numbers
 That is, the graphs for 𝑓 𝑡 = 𝑎𝑡 and 𝑓 𝑡 = 𝑎−𝑡 are always
above the t-axis
Rules of exponential functions
Statement
1. To multiply numbers with the same base –
just ADD the indices
Equation
Example illustrating rule
𝑎𝑚 𝑥𝑎𝑛 = 𝑎𝑚+𝑛
2. To divide numbers with the same base,
SUBTRACT the indices
𝑎𝑚
= 𝑎𝑚−𝑛
𝑛
𝑎
3. To raise an exponential to a power,
MULTIPLY the indices
𝑎𝑚
𝑘
= 𝑎𝑚𝑥𝑘
Rules of exponential functions contd…
Statement
4. To multiply numbers with the same index –
Multiply the bases & raise to the index
5. To divide numbers with the same index,
Equation
Example illustrating rule
𝑎𝑛 𝑥𝑏𝑛 = (𝑎. 𝑏)𝑛
𝑎𝑛
𝑎
=
𝑏𝑛
𝑏
𝑛
Rules of exponential functions contd….
1. Anything raised to the power zero is one, 𝑎0 = 1
2. If an exponential in a fraction is moved from below
the division line,
0
1
𝑎
the sign of the index (power) changes; 𝑛 = 𝑛 = 𝑎0−𝑛 = 𝑎−𝑛 , e.g.,
𝑎
𝑎
1
20
0−4 = 2−4
=
=
2
24
24
3. To write roots as indices, everything under the root sign is raised to
1
3
𝑛
1
𝑛
the power of
: 𝑎 = 𝑎 ; 4 = 41 3 ;
𝑟𝑜𝑜𝑡
4. If any product or fraction is raised to a power, the index on each
term is multiplied by the power:
2𝑏 2
𝑐5
2
=
4𝑏 4
𝑐 10
5. There are no rules for simplifying sums or differences of exponential
functions; e.g., 2𝑥 + 2𝑦 ≠ 2𝑥+𝑦
Solution to exponential functions
• Remember: the solution of an equation is the value of the variable for
which the LHS = RHS
• Solving exponential functions involves two steps:
1. Simplify the exponential equation by writing each side of the
equation as 𝑏𝑎𝑠𝑒 𝑝𝑜𝑤𝑒𝑟
2. Since the bases on each side of the equation are identical, then if
the LHS is to be equal to the RHS, the indices must also be identical.
So equate the indices
3. Examples:
1. 2𝑥+3 2𝑥 = 32
Applications of exponential functions
Growth:
Finding the future value of a sum to be invested in the current period
𝐹 = 𝑃 1 + 𝑖 𝑡 ; Principal (P) compounded annually for t years
𝑖 𝑚𝑡
𝐹 =𝑃 1+
; Principal compounded (m) times a year for t years
𝑚
𝐹 = 𝑃𝑒 𝑟𝑡 ; Principal compounded continuously at interest r, where i, r and t are
constants as in investment and some models of population growth
Discounting:
Finding the present value of a sum to be received in the future
𝑃=
𝐹
1+𝑖 𝑡
=𝐹 1+𝑖
−𝑡
; Present value (P) of F to be received in t years under
annual compounding
𝑖 −𝑚𝑡
𝑃 =𝐹 1+
; Present value (P) of F to be received in t years under multiple
𝑚
compounding
𝑃 = 𝐹𝑒 −𝑟𝑡 ; Present value (P) of F to be received in t years under continuous
compounding
logarithms
 Logarithm is the power to which a base (b) must be raised to obtain a
particular number (y), generally;
 𝑦 = 𝑏 𝑡 ↔ 𝑡 = 𝑙𝑜𝑔𝑏 𝑦
 Log of y to the base b is the power to which the base (b) must be
raised in order to attain the value y
Solving natural exponential & logarithmic functions
 Since they are inverses of each other, one is helpful in
solving the other
Natural exponential
Natural logarithm
Constant
𝑒 ln 𝑎 = 𝑎
ln 𝑒 𝑎 = 𝑎
Variable
𝑒 ln 𝑥 = 𝑥
ln 𝑒 𝑥 = 𝑥
𝑒 ln 𝑓(𝑥) = 𝑓(𝑥)
ln 𝑒 𝑓(𝑥) = 𝑓(𝑥)
Function of a variable
𝐥𝐧 𝒆𝒏 = n
logarithms
 Common log and natural log
 Two commonly used bases – the number 10 & number e
 When 10 is the base, the logarithm is called common
logarithm (𝑙𝑜𝑔10 )
 When e is the base, the logarithm is called natural
logarithm (𝑙𝑜𝑔𝑒 ) or ln
 In analytical work, natural logs are more convenient than
common logs
Rules of logarithms
Statement
Equation
Rule 1. Log of a
product
ln 𝑢𝑣 = ln 𝑢 + ln 𝑣
Rule 2. Log of
quotient
ln 𝑢 𝑣 = ln 𝑢 − 𝑙𝑛𝑣
Rule 3. Log of power
ln 𝑢𝑎 = 𝑎 ln 𝑢 ; 𝑢 > 0
Rule 1, Rule 2 & Rule 3
Rule 4. conversion of
log base
Rule 5. Inversion of
log base
Example illustrating rule
ln 𝑢𝑣 𝑎 /𝑡 = ln 𝑢 + ln 𝑣 𝑎 − ln 𝑡 = ln 𝑢 + 𝑎 𝑙𝑛 𝑣
− ln t
𝑙𝑜𝑔𝑏 𝑢 = 𝑙𝑜𝑔𝑏 𝑒 𝑙𝑜𝑔𝑒 𝑢 ; (𝑢 > 0)
𝑙𝑜𝑔𝑏 𝑒 =
1
𝑙𝑜𝑔𝑒 𝑏
Proof: Let 𝑢 = 𝑒 𝑝 , 𝑠𝑜 𝑡𝑕𝑎𝑡 𝑝 = 𝑙𝑜𝑔𝑒 𝑢, then
𝑙𝑜𝑔𝑏 𝑢 = 𝑙𝑜𝑔𝑏 𝑒 𝑝 = 𝑝 𝑙𝑜𝑔𝑏 𝑒 = (𝑙𝑜𝑔𝑒 𝑢)(𝑙𝑜𝑔𝑏 𝑒)
𝑙𝑒𝑡 𝑢 = 𝑏; 𝑙𝑜𝑔𝑏 𝑏 = (𝑙𝑜𝑔𝑏 𝑒)(𝑙𝑜𝑔𝑒 𝑏); since
𝑙𝑜𝑔𝑏 𝑏 = 1, then 𝑙𝑜𝑔𝑏 𝑒 &(𝑙𝑜𝑔𝑒 𝑏) must be
reciprocal of each other (your practice)
Properties of logs
• There are no real logs of negative numbers
• Logs of numbers less than 1 are negative
• Log (1) = 0 for any base
• Logs of numbers greater than one are always positive
Logarithmic functions
 Logarithmic function refers to a situation where a variable is
expressed as a function of the logarithm of another variable
 Consider: 𝑦 = 𝑏 𝑡 and 𝑦 = 𝑒 𝑡
 Write the above exponential functions in log forms
 In general the log of a function may be written as:
 The base of the power becomes the base of the log,
 The power drops down
 Where the number & base must be real positive numbers
 𝑛𝑢𝑚𝑏𝑒𝑟
 𝑛𝑢𝑚𝑏𝑒𝑟
= 𝑏𝑎𝑠𝑒 𝑝𝑜𝑤𝑒𝑟
= 𝑏𝑎𝑠𝑒 𝑝𝑜𝑤𝑒𝑟
 𝑙𝑜𝑔𝑏𝑎𝑠𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 =
𝑝𝑜𝑤𝑒𝑟
Base conversion
𝑡
• 𝑦 = 𝐴𝑏𝑟𝑡
can always be converted into natural exponential function:
𝑦 = 𝐴𝑒
• Consider a more general expression: 𝐴𝑏 𝑐𝑡 & convert it to 𝐴𝑒 𝑟𝑡
• Idea is to express r as a function of b and c
• We accomplish this by taking natural logs on both sides
• ln 𝑒 𝑟 = ln 𝑏 𝑐
• 𝑟 = ln 𝑏 𝑐 = 𝑐 ln 𝑏
• Therefore y = 𝐴𝑏 𝑐𝑡 can always be rewritten in the natural base form,
• 𝑦 = 𝐴𝑒 (𝑐 ln 𝑏)𝑡
• Convert 𝑦 = 3(5)2𝑡 to a natural exponential function
• A=3; b=5 & c=2 hence 𝑟 = 𝑐 𝑙𝑛 𝑏 = 2 ln 5
• 𝑦 = 3(5)𝑡 = 3𝑒 (2 ln 5)𝑡
• Use your calculate to simplify
Base conversion
• Convert log function of the form, t = 𝑙𝑜𝑔𝑏 𝑦 into equivalent
natural log function
• 𝑙𝑜𝑔𝑏 𝑦 = (𝑙𝑜𝑔𝑏 𝑒)(𝑙𝑜𝑔𝑒 𝑦)
• By rule 5 of logs, t = 𝑙𝑜𝑔𝑏 𝑦 =
•t=
1
𝑙𝑜𝑔𝑒 𝑦
𝑙𝑜𝑔𝑒 𝑏
ln 𝑦
ln 𝑏
• Convert the following functions into the natural log forms
1. t = 𝑙𝑜𝑔2 𝑦
2. t = 7𝑙𝑜𝑔10 2𝑦
Rules of differentiation
• 1. The natural exponential function rule
The derivative of a natural exponential function is equal to
the original exponential function times the derivative of the
exponent
• 𝑦 = 𝑒 𝑔(𝑡)
• 𝑦 ′ = 𝑒 𝑔(𝑡) ∙ 𝑔′ (𝑡)
• Example 1: 𝑦 = 𝑒 𝑘𝑡−𝑐 ; exponent = kt-c
• 𝑔 𝑡 = 𝑘𝑡 − 𝑐 & 𝑔′ 𝑡 = 𝑘
• Substituting: 𝑦 ′ = 𝑒 𝑘𝑡−𝑐 ∙ 𝑘 = 𝑘𝑒 𝑘𝑡−𝑐
Rules of differentiation
• 2. The natural logarithmic function rule
𝑦 = ln 𝑔(𝑡)
•𝑦
•
•
•
•
•
•
1
𝑔′ (𝑡)
′
=
. 𝑔 (𝑡) =
𝑔(𝑡)
𝑔(𝑡)
𝑑
1
= ln 𝑎𝑡 → (ln 𝑎𝑡) = . 𝑎
𝑑𝑡
𝑎𝑡
′
𝑦
Example 1: 𝑦 = ln 5𝑡
𝑔 𝑡 = 5𝑡 & 𝑔′ 𝑡 = 5
5
1
′
Substituting: 𝑦 = =
5𝑡
𝑡
5
Example 2: 𝑦 = ln 𝑡 ;
𝑔 𝑡 = 𝑡 5 & 𝑔′ 𝑡 = 5𝑡 4
′
• Substituting: 𝑦 =
5𝑡 4
𝑡5
=
5
𝑡
=
𝑎
𝑎𝑡
=
1
𝑡
Rules of differentiation
• 3. Exponential function rule for base b other than e
The derivative is the original function times the derivative of the
exponent times the natural log of the base
• 𝑦 = 𝑏 𝑔(𝑡) ; 𝑏 > 0, 𝑏 ≠ 1
•
•
•
•
•
•
•
•
𝑦 ′ = 𝑏 𝑔(𝑡) ∙ 𝑔′ (𝑡) ∙ ln 𝑏
Example 3: 𝑦 = 𝑏 𝑡 ;
𝑔 𝑡 = 𝑡 & 𝑔′ 𝑡 = 1
Substituting: 𝑦 ′ = 𝑏 𝑡 ∙ 1 ∙ ln 𝑏
𝑦 ′ = 𝑏 𝑡 ln 𝑏
Example 4: 𝑦 = 121−𝑡
𝑏 = 12; 𝑔 𝑡 = 1 − 𝑡; 𝑔′ 𝑡 = −1
𝑦 ′ = − (12)1−𝑡 ln 12
Rules of differentiation
• 4. Logarithmic function rule for base b other than e
𝑦 = 𝑙𝑜𝑔𝑎 𝑔(𝑡); 𝑎 > 0, 𝑎 ≠ 1;
•
𝑦′
=
1
𝑔(𝑡)
∙
𝑔′ (𝑡)
∙ 𝑙𝑜𝑔𝑎 𝑒 =
1
𝑔(𝑡)
∙
𝑔′ (𝑡)
• Example 5: 𝑦 = 𝑙𝑜𝑔2 (𝑡 + 1);
• 𝑔 𝑡 = 𝑡 + 1 & 𝑔′ 𝑡 = 1
′
• Substituting: 𝑦 =
•
𝑦′
=
1
𝑡+1 ln 2
1
𝑡+1
∙ 1 ∙ 𝑙𝑜𝑔2 𝑒
∙
1
ln 𝑎
(Rule 5)
Application
𝑑𝑦
• find
𝑑𝑥
𝑓𝑟𝑜𝑚 𝑦 = 𝑥 𝑎 𝑒 𝑘𝑥−𝑐
• Using derivatives of exponential functions rules
1. Using derivatives of exponential functions rules and
2. Taking the natural log of the function & then
differentiating the natural log function
Application
Practice: find the relative growth rate of sales at t=4, given
S t = 100,000𝑒 0.5 𝑡 ,
1. 𝐺 =
𝑆′
𝑆
=
At t=4; G =
0.5
−0.5
0.5𝑡
0.25𝑡
∗100,000𝑒
0.5
100,000𝑒 0.5𝑡
0.25
4
=
0.25
2
=
0.25𝑡 −0.5
=
0.25
𝑡
= 0.125 = 12.5% or
0.5𝑡 0.5
2. ln 𝑆 = ln 1000,000 + ln 𝑒
= ln 100,000 + 0.5𝑡 0.5 & taking
derivative of the natural log with respect to t
1 𝑑𝑆
𝑑
0.25
0.5
−0.5
𝐺= ∙
=
ln 100,000 + 0.5𝑡
= 0 + 0.25𝑡
=
=
𝑆 𝑑𝑡
𝑑𝑡
4
Optimization of Exponential and Logarithmic
Functions
 Unconstrained Optimization Problems
 Suppose you are given the demand function,
 𝑃 = 12.50𝑒 −0.005𝑄
 Determine the price and quantity at which the total revenue
will be maximized
 𝑇𝑅 = 𝑃𝑄 = (12.50𝑒 −0.005𝑄 )𝑄
 FOC: 𝑇𝑅′ = −0.005(12.50𝑒 −0.005𝑄 )𝑄 + (1) 12.50𝑒 −0.005𝑄 = 0
 12.50𝑒 −0.005𝑄 1 − 0.005𝑄 = 0
 Since 12.50𝑒 −0.005𝑄 ≠ 0 for any value of 𝑄, then
 1 − 0.005𝑄 = 0, 𝑄 = 200
 𝑃 = 12.50𝑒 −0.005(200) = 12.5𝑒 −1 = 4.60
Optimization of Exponential and Logarithmic
Functions
 SOC:
𝑇𝑅′′ = 12.50𝑒 −0.005𝑄 −0.005 + (1 − 0.005𝑄)(−0.005)(12.5𝑒 −0.005𝑄 )
 Evaluated at 𝑄 = 200 , 𝑇𝑅′′ = −0.0625 0.36788 < 0 (Sufficient
condition for maximum satisfied)
Optimization of Exponential and Logarithmic
Functions
 Optimal Timing
 The optimum time is the time over which a good whose value
appreciates overtime is stored so that its value at an approximate
time in future is maximized before it is sold.
 Such goods include cheese, wine etc.
 The most popular function to represent the value of such goods is
the exponential function
 The Problem of Cheese Storage
 Suppose a cheese maker has in store a quantity of cheese worth 𝐴
Pula at time 𝑡 = 0
Optimization of Exponential and Logarithmic
Functions
 The cheese maker has the choice of selling the
cheese at the present time (𝑡 = 0) or storing and
selling it later when its value has appreciated
 Suppose the growing value of the cheese is
represented by the function, 𝑉 = 2 𝑡
 Note: At time 𝑡 = 0, the cheese maker will receive,
𝑉 = 20 = 1 Pula
 As a businessman, the cheese maker will choose to
sell the cheese in future when he can maximize
profit obtained from the sale of the cheese
Optimization of Exponential and Logarithmic
Functions
 Assuming storage costs are zero, his profit will be
equivalent to the revenue received
 The maximization of profits involves maximization of
the present value of 𝑉, where 𝑉 = 2 𝑡 . The present
value of 𝑉 can be denoted 𝑃(𝑡)
 Assuming continuous interest compounding, the
present value of 𝑉 is defined, 𝑃 𝑡 = 𝑉𝑒 −𝑟𝑡
 Recalling that 𝑉 = 2 𝑡 , then 𝑃 𝑡 = (2 𝑡 )𝑒 −𝑟𝑡
 Therefore, 𝑃(𝑡) = 2 𝑡 𝑒 −𝑟𝑡
 (after substituting and simplifying)
Optimization of Exponential and Logarithmic
Functions
 Three steps for finding the optimal time:
i. Express the function, 𝑃(𝑡) in natural log form
ii.
Apply the necessary condition for optimization: find the first order
derivative, set it equal to zero and solve the resulting equation to
determine the critical value for the function
iii. Evaluate if the sufficient condition is satisfied: the sufficient
condition for maximum is satisfied if the second derivative is
negative
Optimization of Exponential and Logarithmic Functions










Step 1: Express the function in natural log form
𝑃(𝑡) = 2 𝑡 𝑒 −𝑟𝑡 , 𝑙𝑛𝑃 = ln 2 𝑡 + ln 𝑒 −𝑟𝑡
𝑙𝑛𝑃 = 𝑡ln 2 − 𝑟𝑡
Step 2: First Order Condition
1
1 𝑑𝑃
𝑑
𝑑
= 𝑙𝑛𝐴 +
𝑡 2 ln 2 − 𝑟𝑡
𝑃 𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑃
1 −1
= 𝑃 𝑡 2 𝑙𝑛2 − 𝑟 = 0
𝑑𝑡
2
𝑑𝑃
ln 2
=𝑃
−𝑟 =0
𝑑𝑡
2 𝑡
ln 2
Since 𝑃 ≠ 0, then
− 𝑟=0
2 𝑡
ln 2
ln 2
= 𝑟 𝑜𝑟 𝑡 =
2 𝑡
2𝑟
2
ln 2
∗
𝑡 =
2𝑟
=0
Optimization of Exponential and Logarithmic
Functions
 Supposing the interest rate is 5 %, then
 𝑟 = 0.05
𝑡∗
0.6931 2
2(0.05)
 Solving for 𝑡, =
= (6.931)2 = 48.0 𝑦𝑒𝑎𝑟𝑠
 48 years is the Optimum time
 Step 3: Evaluate the sufficient condition
𝑑2 𝑃
𝑑 1 −1
1 −1
𝑑𝑃
 2 = 𝑃.
𝑡 2 𝑙𝑛2 − 𝑟 + 𝑡 2 𝑙𝑛2 − 𝑟
,
𝑑𝑡


𝑑𝑡 2
2
𝑑𝑃
Since
= 0 at the critical value,
𝑑𝑡
𝑑2 𝑃
It means, 2 = 𝑃 … . =?
𝑑𝑡
𝑑𝑡
Optimization of Exponential and Logarithmic
Functions
 Practice Problems
1. Cut glass currently worth $100 is appreciating in value according
1
𝑡2
to, 𝑉 = 100𝑒 𝑡 = 100𝑒 . How long should the cut glass be kept to
maximize its present value if under continuous compounding (a)
𝑟 = 0.08 and (b) 𝑟 = 0.12 (Dowling, pp. 194)
2. Land bought for speculation is increasing in value according to
3
the formula, 𝑉 = 1000𝑒 𝑡 . The discount rate under continuous
compounding is 0.09. How long should the land be held to
maximize the present value (Dowling, pp. 195)