Chapter 8-x Applications of Exponential and Logarithmic Functions Day 1
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Transcript of Chapter 8-x Applications of Exponential and Logarithmic Functions Day 1
Essential Question: What are some types of real-life problems where exponential or
logarithmic equations can be used?
Compound Interest◦ A = P(1 + )nt, where:
A = Amount at the end of compounding P = Principal (starting) amount r = Interest rate (as a decimal) n = Number of times per year compounded t = number of years
◦ We’re only going to be dealing with situations where interest is compounded yearly, so we will use the formula: A = P(1 + r)t
r
n
Example◦ A computer valued at $6500 depreciates at the
rate of 14.3% per year.◦ Write a function that models the value of the
computer. A = P(1 + r)t
Do we know A (the value at the end)? Do we know P (the value at the start)? Do we know r (the rate)? Do we know t (the number of years)?
◦ Find the value of the computer after three years.
NoYes, 6500Yes, 0.143No
A = 6500(1 – 0.143)t = 6500(0.857)t
A = 6500(0.857)3 = $4091.25
depreciates
Continuously Compounding Interest◦ So let’s bring “n” back for just one example:◦ Suppose you invest $1 for one year at 100%
annual interest, compounded n times per year. Find the maximum value of the investment in one year.
◦ Observe what happens to the final amount as n grows larger and larger.
Compounding Continuously◦ Annually
◦ Semiannually
◦ Quarterly
◦ Monthly
◦ Daily
◦ Hourly: 365 • 24 = 8760 periods
◦ Every minute: 8760 • 60 = 525,600 periods
◦ Every second: 525,600 • 60 = 31,536,000
periods
1111(1 ) $2.00A
4(1)141(1 ) $2.4414A
2(1)121(1 ) 2.25A
12(1)1121(1 ) $2.6130A
365(1)13651(1 ) $2.71457A
8760(1)187601(1 ) $2.718127A
525600(1)15256001(1 ) $2.7182792A
31536000(1)1315360001(1 ) $2.7182825A
$2.7182825 is the same as the number e to five decimal places (e = 2.71828182…)
So if we’re compounding continuously (instead of yearly), we can instead use the equation◦ A = Pert, where
A = Amount at the end of compounding P = Principal (starting) amount r = Interest rate (as a decimal) t = number of years
Example◦ Suppose you invest $1050 at an annual interest
rate of 5.5% compounded continuously. How much money will you have in the account after 5 years? A = Pert
Do we know A (the value at the end)? Do we know P (the value at the start)? Do we know r (the rate)? Do we know t (the number of years)?
NoYes, 1050Yes, 0.055
Yes, 5
A = 1050 e(0.055 • 5) = $1382.36
Your Turn◦ Suppose you invest $1300 at an annual interest
rate of 4.3% compounded continuously. How much money will you have in the account after 3 years? A = Pert
A = 1300(e)(0.043 • 3) = $1479.00
Assignment◦ Worksheet◦ Round all problems appropriately (if talking about
money: 2 decimal places; if talking about population: nearest integer)
Essential Question: What are some types of real-life problems where exponential or
logarithmic equations can be used?
Radioactive Decay◦ The half-life of a radioactive substance is the time
it takes for half of the material to decay.◦ It’s most often used for things like carbon-14
dating, which determines how old a substance is◦ The function for radioactive decay is
where: P = the initial amount of the substance x = 0 corresponds to the time since decay began h = the half-life of the substance
(0.5)xhy P
Example◦ A hospital prepares a 100-mg supply of
technetium-99m, which has a half-life of 6 hours. Write an exponential function to find the amount of technetium-99m.
◦ Do we know y (the value at the end)? Do we know P (the value at the start)? Do we know x (the amount of time)? Do we know h (the half-life)?
◦ Use your function to determine how much technetium-
99m remains after 75 hours.
(0.5)xhy P
NoYes, 100NoYes, 6
6100(0.5)x
y
756100(0.5) 0.017y mg
Example◦ Arsenic-74 is used to locate brain tumors. It has a
half-life of 17.5 days. Write an exponential decay function for a 90-mg sample.
◦ Use the function to find the amount remaining after 6 days.
17.590(0.5)x
y
617.590(0.5) 71y mg
Loudness◦ Logarithms are used to model sound. The intensity
of a sound is a measure of the energy carried by the sound wave. The greater the intensity of a sound, the louder it seems. This apparent loudness L is measured in decibels. You can use the formula
, where I is the intensity of the sound in watts per square
meter (W/m2) I0 is the lowest-intensity sound that the average human
ear can detect. (We will use I0 = 10-12 w/m2)
0
10 logI
LI
Example◦ Suppose you are the supervisor on a road
construction job. Your team is blasting rock to make way for a roadbed. One explosion has an intensity of 1.65 x 10-2 W/m2. What is the loudness of the sound in decibels?
◦ 0
10 logI
LI
2
12
1.65 1010 log 102
10L db
Assignment◦ Worksheet◦ Round all problems appropriately
For half life, round to two decimal places For loudness, round to the nearest dB
Essential Question: What are some types of real-life problems where exponential or
logarithmic equations can be used?
Acidity◦ Scientists use common logarithms to measure
acidity, which increases as the concentration of hydrogen ions in a substance. The pH of a substance equals:
pH = –log[H+],where [H+] is the concentration of hydrogen ions.
Example The pH of lemon juice is 2.3, while the pH of milk is
6.6. Find the concentration of hydrogen ions in each substance. Which substance is more acidic?
Lemon Juice Milk (Your turn) pH = -log[H+]
2.3 = -log[H+] substitutelog[H+] = -2.3 divide by -1[H+] = 10-2.3 convert to exp[H+] = 5.0 x 10-3
pH = -log[H+]6.6 = -log[H+]log[H+] = -6.6[H+] = 10-6.6
[H+] = 2.5 x 10-7
The lemon juice is more acidic as it contains more hydrogen ions
Compound Interest (finding something other than A)◦ Remember our formulas from Monday:
A = P(1 + r)t for compounding annually
A = Pert for compounding continuously
◦ Finding P How much should be invested at 5.5% compounded
annually to yield $3500 at the end of 4 years? 3500 = P(1 + 0.055)4
3500 = P(1.055)4
3500/(1.055)4 = P$2825.26= P
Compound Interest (finding something other than A)◦ Finding r
What interest rate would be required to grow an investment of $1000 to $1407.10 in seven years if that interest is compounded annually?
1407.10 = 1000(1 + r)7
1.4071 = (1 + r)7
1.05 = 1 + r0.05 = r, meaning an interest rate of 5%
Compound Interest (finding something other than A)◦ Finding t
How long will it take to double an investment of $500 at 7% interest, compounded annually?
1000 = 500(1 + 0.07)t
2 = (1.07)t
log1.07 2 = t
10.25 = t, meaning 10.25 years
log 2/log 1.07 = t
Assignment◦ Worksheet◦ Round all problems appropriately
If talking about money: 2 decimal places