Calculus for Business
Lesson- properties, equations with exponents and
power and exponential functions



Objectives:
Name:____________________________________
Date:_____________________________________
use the properties of exponents
solve equations containing rational exponents
examine power and exponential functions
Do Now: Use the exponential properties to simplify and rewrite the following expressions:
(1)
ax  ay  a x y
(2)
a 
(3)
abx  a x b y
(4)
ax
a
   x
b
b
(5)
ax
 a x y
y
a
(6)
a x 
(7)
a0  1
x y
 a xy
x
1
ax
__________________________________________________________________________________________
Using Exponential Function Properties to Solve for x:
Process 1
~get bases =
~set exponents =
~solve
Process 2
~root or power on both sides
~solve
Examples (each relates to “Process 1”):
1.
44 x1  42 x2
2.
45 x1  16 2 x1
3.
3x  9 x4
2
3 x  (32 ) x  4
2
4x  1  2x  2
2 x  3
3
x
2
4 5 x 1  (4 2 ) 2 x 1
3 x  3 2 x 8
4 5 x 1  4 4 x 2
5x  1  4 x  2
x 2  2x  8
x  3
( x  4)( x  2)  0
X  4,2
2
x 2  2x  8  0
More Examples (each relates to “Process 2”):
x 4  81
4.
5.
x 1  4
x 4  4 81
x  3
4


2
x 1  4
x  1  16
6.
(2 x  1)5  32
5
2
x  17
(2 x  1) 5  5  32
2 x  1  2
2 x  1
1
x
2
Power function:
y  a  xb
exponential function:
y  a bx
Small Group Activity
On your graphing calculator, simultaneously graph: y = 0.5x, y = 0.75x, y = 2x, y = 5x
(1)
What is the range of each exponential function? 0,  
(2)
What is the behavior of each graph? Horizontal asymptote on x-axis, increasing if b>1, decreasing if b<1
(3)
Do the graphs have any asymptotes? y=0
(4)
(a) What point is on the graph of each function? (0,1)
(b) Why? Anything to the 0th power is 1.
Characteristics of graphs of y = nx
n>1
0<n<1
 , 
 , 
range
0, 
0, 
y-intercept
(0, 1)
(0, 1)
Increasing
decreasing
y=0
y=0
None
None
domain
behavior
horizontal asymptote
vertical asymptote
2
Graphing exponential functions, exponential growth and decay
Objectives:



graph exponential functions
use exponential functions to determine growth and decay
half-life
Using Exponential Functions for Real World Applications:
Exponential Growth or Decay: N = N0 (1 + r)t
N= Final Amount
N0=Initial Amount r = rate of growth (+) or decay ( – ) t = time
__________________________________________________________________________________________
(1) A researcher estimates that the initial population of honeybees in a colony is 500. They are increasing at a
rate of 14% per week. What is the expected population in 22 weeks?
N  500(1  0.14) 22  8930.519719
(2) Write a formula that represents the average growth of the population of a city with a rate of 7.5% per year.
Let x represent the number of years, y represent the most recent total population of the city, and A is the
city’s population now. What is the expected population in 10 years if the city’s population now is 22,750
people? Graph the function for 0  x  20.
N  A(1  0.075) t
Graph:
y  22750(1  0.075) t  22750(1.075) t
N  22750(1  0.075)10  46888.46804
__________________________________________________________________________________________
(3) Mexico has a population of about 100 million people, and it is estimated that the population will double in
21 years. If population growth continues at the same rate, what will be the population in:
(a) 15 years
(b) 30 years
(c) graph the population growth for 0  time  50
2  1(1  r ) 21
N  100(1  .0336) t
2  (1  r ) 21
N  100(1  .0336)15
N  164.167229 mill
21
2  21 (1  r ) 21
21
2  1 r
21
2 1  r
.033557783  r
Graph:
y  100(1  0.0336) t  22750(1.0336) t
N  100(1  .0336) 30
N  269.5100841mill
3
(4) In 1990, Exponential City had a population of 700,000 people. The average yearly rate of growth is 5.9%.
Find the projected population for 2010.
N  700000(1  0.059) 20  2203013.924
(5) Find the projected population of each location in 2015:
(a)
In Honolulu, Hawaii, the population was 836,231 in 1990. The average yearly rate of growth is
0.7%.
N  836231(1  0.07) 25  4538587.424
(b)
The population in Kings County, New York has demonstrated an average decrease of 0.45% over
several years. The population in 1997 was 2,240,384.
N  2240384(1  0.0045)18  2065690.351
(6) The population of Los Angeles County was 9,145,219 in 1997. If the average growth rate is 0.45%, predict
the population in 2010.
N  9145219(1  0.0045)13  9694900.229
Graph the equation for 0  time  20.
(7) Radioactive gold 198 (198Au), used in imaging the structure of the liver, has a half-life of 2.67 days. If the
initial amount is 50 milligrams of the isotope, how many milligrams (rounded to the nearest tenth) will be
left over after:
(a) ½ day
(b) 1 week
1
 1(1  r ) 2.67
2
1 2.67
2.67
 (1  r ) 2.67
2
2.67
N  50(1  .229) 0.5
N  43.903
N  50(1  .229) 7
N  8.0975
1
 1  r
2
1
r
2
.2286643175  r
1  2.67
4
(8) If a farmer uses 25 pounds of insecticide, assuming its half-life is 12 years, how many pounds (rounded to
the nearest tenth) will still be active after:
(a) 5 years
(b) 20 years
1
 1(1  r )12
2
1 12
12
 (1  r )12
2
12
N  25(1  .0561) 5
N  18.731
N  50(1  .229) 20
N  7.879
1
 1  r
2
1
r
2
.0561256873  r
1  12
(9)
In 2000, the chicken population on a farm was 10,000. The number of chickens increased at a rate of 9%
per year. Predict the population in 2005.
N  10000(1  0.09) 5  15386.23955
Graph the equation for 0  time  15.
(10)If Kenya has a population of about 30,000,000 people and a doubling time of 19 years and if the growth
continues at the same rate, find the population (rounded to the nearest million) in:
(a) 10 years
(b) 30 years
2  1(1  r )19
N  30(1  .0372) t
2  (1  r )19
N  30(1  .0372)10
19
2  19 (1  r )19
19
2  1 r
2 1  r
.0371550444  r
19
N  43.22612805 mill
N  30(1  .0372) 30
N  89.74215572 mill
5
Compound Interest
Objectives:

use exponential functions to determine compound interest
(1) Rates can be compounded in different increments per year. Exponential growth occurs how often if the
rate is compounded:
annually:
1
bi-annually:
2
quarterly:
4
monthly:
12
weekly:
52
daily:
365
The general equation for exponential growth is modified for finding the balance in an account that earns
compound interest.
nt
r

Compound Interest: A  P1  
n

A = Final Amount
P = Initial Amount (Principle Balance)
r = rate
n = number of compounding periods
t = time in years
_________________________________________________________________________________________
(1) If Charlie invested $1,000 in an account paying 10% compounded monthly, how much will be in the
account at the end of 10 years?
1012
 .10 
A  10001 

 12 
 $2707.04
(2) Mike would like to have $20,000 cash for a new car 5 years from now. How much should be placed in an
account now if the account pays 9.75% compounded weekly?
 .0975 
20000  A1 

52 

525
 $12288.81
(3) Suppose $2,500 is invested at 7% compounded quarterly. How much money will be in the account in:
(a) ¾ year
(b) 15 years
 0.07 
A  25001 

4 

0.754
 $2633.56
154
 0.07 
A  25001 

4 

 $7079.54
6
_________________________________________________________________________________________
(4) Suppose $4,000 is invested at 11% compounded weekly. How much money will be in the account in:
(a) ½ year
(b) 10 years
 0.11 
A  40001 

52 

0.552
 $4225.92
1052
 0.11 
A  40001 

52 

 $12002.71
(5) How much money must Cindy invest for a new yacht if she wants to have $50,000 in her account that earns
5% compounded quarterly after 15 years?
 .05 
50000  A1 

4 

415
 $23728.38
(6) Carol won $5,000 in a raffle. She would like to invest her winnings in a money market account that
provides an APR of 6% compounded quarterly. Does she have to invest all of it in order to have $9,000 in
the account at the end of 10 years? Show your work and explain your answer.
 .06 
9000  A1 

4 

410
 $4961.36
No. She only needs to invest $4961.36
Exponential Functions with base e

Objective:
use exponential functions with base e
Euler Savings Bank provides a savings account that earns compounded interest at a rate of 100%. You may
choose how often to compound the interest, but you can only invest $1 over the course of one year.
 1
y  11  
 x
x
The limit as x gets farther and farther away from 0
appears to be Euler’s number which is approximately
2.718 or e.
This means that as the number of compounding
periods increases, the growth rate seems to maximize
itself at Euler’s number (e).
7
Exponential Growth or Decay (in terms of e): N = N0 ekt
N= Final Amount
(1)
N0=Initial Amount
k = rate of growth (+) or decay ( – ) t = time
According to Newton, a beaker of liquid cools exponentially when removed from a source of heat.
Assume that the initial temperature Ti is 90F and that k = 0.275.
(a) Write a function to model the rate at which the liquid cools.
N  90e .275t
(b) Find the temperature T of the liquid after 4 minutes (t)
N  90e .2754  29.95839753
(c) Graph the function and use the graph to verify your answer in part (b)
(2)
Suppose a certain type of bacteria reproduces according to the model: B  100e 0.271t , where t is the time in
hours.
(a) At what percentage rate does this type of bacteria reproduce?
(b) What was the initial number of bacteria?
27.1%
100
(c) Find the number of bacteria (rounded to the nearest whole number) after:
(i) 5 hours
388
(ii) 1 day
66781
(iii) 3 days
29782004910
(3)
A city’s population can be modeled by the equation y = 33,430e0.0397 t , where t is the number of years
since 1950.
(a) Has the city experienced a growth or decline in population?
(b) What was the population in 1950?
(c) Find the projected population in 2010.
growth
33430
361931.0414
8
(4) If you invest $5,250 in an account paying 11.38% compounded continuously, how much money will be in
the account at the end of:
N  5250e.11386.25  $10691.81
(a) 6 years 3 months
N  5250e.113817  $36336.70
(b) 204 months
__________________________________________________________________________________________
(5) A promissory note will pay $30,000 at maturity 10 years from now. How much should you be willing to
pay for the note now if the note gains value at a rate of 9% compounded continuously?
30000  N 0 e 0.09*10
$12197.09  N 0
Continuous Compound Interest
Objective:

use exponential functions to determine continuously compounded interest
Continuously Compounded Interest: A = Pert
A = Final Amount
P = Initial Amount (Principle Balance)
r = rate
t = time in years
(1) Tim and Kerry are saving for their daughter’s college education. If they deposit $12,000 in an account
bearing 6.4% interest compounded continuously, how much will be in the account when she goes to college
in 12 years?
A  12000e.06412  $25865.41
(2) Paul invested a sum of money in a certificate of deposit that earns 8% interest compounded continuously.
If Paul made the investment on January 1, 1995, and the account was worth $12,000 on January 1, 1999,
what was the original amount in the account?
12000  Pe.084  $8713.79
(3) Compare the balance after 30 years of a $15,000 investment earning 12% interest compounded
continuously to the same investment compounded quarterly.
A  15000e
.1230
 $548973.52
 .12 
A  150001 

4 

Difference of $28308.71
430
 $520664.81
(4) Given the original principal, the annual interest rate, the amount of time for each investment, and the type
of compounded interest, find the amount at the end of the investment:
(a) P = $1,250;
r = 8.5%;
t = 3 years;
 0.085 
A  12501 

2 

(b) P = $2,575;
compounded semi-annually
2*3
 $1604.60
r = 6.25%;
t = 5 years 3 months;
0.0625*5.25
A  2575e
 $3575.03
compounded continuously
9
Properties of a logs, rewriting Exponential functions as logarithms, log graphs
Objective:
 To learn what a logarithm is
 To learn the properties of logs
 To learn to rewrite an exponential function as a logarithm
 Graphing logs
_________________________________________________________________________________________
What is a logarithm?
logarithm
1610s, Mod.L. logarithmus, coined by Scottish mathematician John Napier (1550-1617), lit. "ratio-number," from Gk.
logos "proportion, ratio, word" (see logos) + arithmos "number" (see arithmetic). Related: Logarithmic.
exponent
1706, from L. exponentem (nom. exponens), prp. of exponere "put forth" (see expound). A mathematical term at first; the
sense of "one who expounds" is 1812. As an adjective, from 1580s.
Logarithms are exponents. In some ways they can be inverses of exponential functions. Logarithms are functions
because exponential functions are one-to-one functions.
We cannot solve an equation like: y  2 x using the algebraic techniques we have learned so far. Therefore, we must try
an alternative technique.
Rule: x  b y is equivalent to y  log b x
The log to the base b is the exponent to which b must be raised to obtain x.
Properties of Logs
logb 1  0
logb b  1
logb b x  x
blog b x  x , where x > 0
logb MN  logb M  logb N
M
logb  logb M  logb N
N
logb Mp  p logb M
Example:
Convert each into logarithmic form
1. y  2 x
log 2 y  x
2. 3  9
39
1
2
log 9 3 
3.
1
 51
5
Convert each into logarithmic form
1
251 / 2  5
4. log 25 5 
2
5. log a b  c
ac  b
1
6. log 3    2
9
32 
1
2
log 5 1  1
5
1
9
10
What is a Natural Logarithm? Log with base e
Rule: x  b y is equivalent to y  log b x
The log to the base b is the exponent to which b must be raised to obtain x.
Properties of Logs
ln 1  0
ln b  1
ln e x  x
e ln x  x , where x > 0
ln MN  ln M  ln N
M
ln
 ln M  ln N
N
ln M p  p ln M
Example:
Convert each into logarithmic form
log e y  x
1. y  e x
ln y  x
Convert each into logarithmic form
ex  5
4. ln 5  x
2. e  x
5. ln b  c
ec  b
6. ln y  2
e 2  y
e  x1 / 2
log x e 
3.
1
 e 1
e
1
2
log e 1 e  1
Example:
Graph each of the following on the same set of axes using the graphing calculator.
1. y  2 x
2. x  2 y
3. log 2 y  x
4. log 2 x  y
5. y  e x
6. x  e y
7. ln y  x
8. ln x  y
11
Simplify log expressions, common logs, evaluate



Objectives:
simplify expressions using the properties of logarithmic functions
define common logarithms
evaluate expressions involving logarithms
Problem Set: write the following expressions in simpler logarithmic forms:
(1)
(3)
logb u2 v 7 = 2 log b u  7 log b v
log b
m
n
2
3
1
2
=
2
1
log b m  log b n
3
2
(2)
logb
1
=  2 log b a
a2
(4)
log b
u
= log b u  log b v  log b w
vw
(6)
logb
3
1
log b x
2
1
n
= log b n  2 log b p  3 log b q
3
3
pq
(5)
logb x =
(7)
Use logarithmic properties to find the value of x (without using a calculator):
1
2
3 4
log b x  log b 9  log b 8  log b 6
log b x  log b
2
3
6
1/ 2
2/3
log b x  log b 9  log b 8  log b 6
log b x  log b 2
log b x  log b 3  log b 4  log b 6
2
x2
Write each expression in terms of a single logarithm with a coefficient of one:
(8)
5 logb x  4 logb y = log b x 5 y 4
(10) 3 logb x  2 logb y 
1
x3 y 2
logb z = log b
4
4
z
3
w2 / 3
(12) logb w  2 logb u = log b 2
2
u
(9)
ie : 2 log b u  log b v  log b
2 logb x  logb y = log b
u2
v
x2
y
(11)  8 logb c = log b c 8
(13)
1/ 3
1
logb (a 2  b 3 ) = log b a 2  b 3 
3
Common Logarithm:
log 10  log x
Change of Base Formula:
log b a 
log a ln a log p a


log b ln b log p b
12
Given loga n, evaluate each logarithm to four decimal places:
(14) log8 172 = 2.4754
(15) log 6 1.258 = .1281
(16) log13 0.0065 = -1.9637
Properties of Logarithmic Functions, Simplifying logarithmic expressions
Objectives:


examine properties of logarithmic functions
simplify expressions using the properties of logarithmic functions
Use the properties of logarithmic functions to solve for x:
(1)
log5 x  2
5 2  x 
(3)
logx 8  3
x3  8
x2
1
25
(2)
log4 64  x
(4)
log8 x 
3
2
3
82 / 3  x  4
Use the properties of logarithmic functions to simplify each expression:
(5)
log8 8
1
(6)
log0.5 1
0
(7)
log10 1,000
3
(8)
log2 64
6
(9)
log7 343
3
(10)
log10 0.001
-3
(11)
loge e
1/2
(12)
log5 3 5
1/3
Write the following expressions in simpler logarithmic forms:
(13)
logb x 6 y 9 = 6 log b x  9 log b y
(15)
logb
(17)
logb 5 x =
mn
= log b m  log b n  log b p  log b q
pq
1
log b x
5
(14)
v7
logb 8 = 7 log b v  8 log b u
u
(16)
logb
(18)
logb 3 x 2  y 2 =
1
=  4 log b a
a4
1
log b ( x  y)  log b ( x  y)
2
13
Application Questions
Objectives:
(1)

solve real-world applications with natural logarithmic functions
Ana is trying to save for a new house. How many years, to the nearest year, will it take Ana to triple the
money in her account if it is invested at 7% compounded annually?
 0.07 
3  11 

1 

1*t
3  1.07 
t
ln 3  ln 1.07 
t
ln 3  t ln 1.07
ln 3
 t  16
ln 1.07 
(2)
At what annual percentage rate (to the nearest hundredth of a percent) compounded continuously will
$6,000 have to be invested to amount to $11,000 in 8 years.
11
11
1 11
11000
11
11000  6000e 8 r
 e8r
 e 8r
ln  ln e 8 r
ln  8r ln e
ln  r  7.58%
6000
6
6
6
8 6
_________________________________________________________________________________________
(3) In 1990, Exponential City had a population of 142,000 people. In what year will the city have a
population of about 200,000 people if it was growing at an exponential rate of k = 0.014?
200000  142000e 0.014t
(4)
200000 200
200

 e 0.014t
ln
 ln e 0.014t
142000 142
142
200
ln
142  t  24.464  1990  2014.464
0.014
In the year 2015.
ln
200
 0.014t ln e
142
If $5,000 is invested at an annual interest rate of 5% compounded quarterly, how long will it take the
investment to double?
 0.05 
2  11 

4 

4*t
 0.05 
ln 2  ln 1 

4 

4*t
ln 2  4t ln 1.0125
ln 2
 t  13.949 years
4 ln 1.0125
(5)
What was the annual interest rate (to the nearest hundredth of a percent) of an account that took 12 years
to double if the interest was compounded continuously and no deposits or withdrawals were made during
the 12-year period?
ln 2
2  1e12r
ln 2  ln e12r
ln 2  12r
 r  0.0578  5.78%
ln 2  12r ln e
12
(6) If a car originally costs $18,000 and the average rate of depreciation is 30%, find the value of the car to the
nearest dollar after 6 years.
N  18000(1  0.3) 6  $2117.68
(7) How many years, to the nearest year, will it take for the balance of an account to double if it is gaining 6%
interest compounded semiannually?
 0.06 
2  11 

2 

2*t
2  1.03 2t
ln 2  ln 1.032t
ln 2  2t ln 1.03
ln 2
 t  11.725  12 years
2 ln 1.03
14
(8) When Rachel was born, her mother invested $5,000 in an account that compounded 4% interest monthly.
Determine the value of this investment when Rachel is 25 years old.
12*25
 0.04 
A  50001 
 $13568.83

12 

_________________________________________________________________________________________
(9) The decay of carbon-14 can be described by the formula A  A 0 e 0.000124 t . Using this formula, how many
years, to the nearest year, will it take for carbon-14 to diminish to 1% of the original amount?
0.01  1e .000124t
ln 0.01  ln e .000124t
ln 0.01  0.000124t ln e
ln 0.01  0.000124t
ln 0.01
 t  37138.46924  37138 years
 0.000124
(10)
In 2002, a farmer had 400 pigs on his farm. He estimated that this population of pigs will double in 15
years. If population growth continues at the same rate, predict the number of pigs in:
a. 2010
b. 2030
2  1(1  r )15
N  400(1  .0473) t
2  (1  r )15
N  400(1  .0473) 8
15
2  15 (1  r )15
15
2  1 r
N  578.9336851 mill
N  400(1  .0473) 28
2 1  r
.0472941228  r
15
(11)
N  1458.985213 mill
If the world population is about 6 billion people now and if the population grows continuously at an
annual rate of 1.7%, what will the population be (to the nearest billion) in 10 years from now?
N  6e 0.017*10  7.111829108  7 billion
__________________________________________________________________________________________
(12) If $100 is invested in an account that has an interest of 7% compounded quarterly, how long will it take
for the balance to reach a value of $1,000?
4t
 0.07 
4t
10  1.0175
1000  1001 

4 

ln 10
 t  33.18106367
4 ln 1.0175
(13)
4t
ln 10  4t ln 1.0175
What interest rate (to the nearest hundredth of a percent) compounded monthly is required for an $8,500
investment to triple in 5 years?
r 

3  11  
 12 

ln 10  ln 1.0175
12*5
r 

3  1  
 12 
60
60
3
60
r 

1  
 12 
60
60
3  1
r
12

12 60 3  1  r  0.2217463732  22.17%
15
(14)
An optical instrument is required to observe stars beyond the sixth magnitude, the limit of ordinary
vision. However, even optical instruments have their limitations. The limiting magnitude L of any
optical telescope with lens diameter D, in inches, is given by the equation L  8.8  5.1 log D . Use this
equation to find the following to the nearest tenth:
a.
the limiting magnitude for a homemade 6-inch reflecting telescope.
L  8.8  5.1log D
b.
L  8.8  5.1log 6
L  12.76857138  12.8
the diameter of a lens that would have a limiting magnitude of 20.6.
20.6  8.8  5.1log D
11.8  5.1log D
11.8
 log D
5.1
10
11.8
5.1
 D  205.9327837
Review- Exponential and Logarithmic Functions
Write each expression in terms of simpler logarithmic forms:
4
(1)
5
log b x y
5 log b x  log b y
(2)
s5
log b 7
u
(3)
log b
1
c8
(4)
 8 log b c
4
log b s  7 log b u
5
m 5n 3
logb
p
5 log b m  3 log b n  0.5 log b p
Given loga n, evaluate each logarithm to four decimal places:
(5)
log3 42 = 3.4022
(6)
log 1 5 = -2.3219
(7)
log6 0.00098 = -3.8666
2
Solve each equation and round answers to four decimal places where necessary:
(8)
log2 x  3
1/8
(9)
log5 4  log5 x  log5 36
log 5 4 x  log 5 36
4 x  36
x9
(10) 1000  75e0.5 x
(11) log 6 x  2
1000
 e 0.5 x
75
1000
ln
 ln e 0.5 x
75
1000
ln
 0. 5 x
75
x  5.181
16
(12) log 7
1
x
49
(14) 10 x  27.5
ln 10 x  ln 27.5
x
ln 27.5
 1.439
ln 10
-2
(13) log x 4 
1
2
16
(15) log x  log 5  log 2  log( x  3)
x
2
 log
5
x3
x
2

5 x3
x( x  3)  10
log
x 2  3 x  10  0
( x  5)( x  2)  0
x5
(16) log x  log 2  1
(17) log4 x  3
64
log 2 x  1
101  2 x
x5
(18) log9 (5  x )  3 log9 2
log 9 (5  x)  log 9 8
5 x 8
x  3
(20) 2  1.002 4 x
20
1
x
20
101 
x
x2
log
(21) e 25 x  1.25
ln 2  ln 1.002 4 x
ln e 25 x  ln 1.25
ln 2  4 x ln 1.002
ln 2
x
 86.73
4 ln 1.002
25 x ln e  ln 1.25
ln 1.25
x
 0.00893
25
(22) log( x  10)  log( x  5)  2
log( x  10)( x  5)  2
(23) log 6 216 
1
log 6 36  log 6 x
2
10 2  ( x  10)( x  5)
100  ( x  10)( x  5)
log 6 216  log 6 361 / 2  log 6 x
100  x 2  5 x  50
216
 log 6 x
6
36  x
0  x 2  5 x  150
x  15x  10  0
x  10
(24)
(19) log 20  log x  1
log 6 216  log 6 6  log 6 x
log 6
Anthony is an actuary working for a corporate pension fund. He needs to have $14.6 million grow to
$22 million in 6 years. What interest rate (to the nearest hundredth of a percent) compounded annually
does he need for this investment?
17
(25)
a.
b.
c.
(26)
The number of guppies living in Logarithm Lake doubles every day. If there are four guppies initially:
Express the number of guppies as a function of the time t.
Use your answer from part (a) to find how many guppies are present after 1 week?
Use your answer from part (a) to find, to the nearest day, when will there be 2,000 guppies?
The relationship between intensity, i, of light (in lumens) at a depth of x feet in Lake Erie is given by
i
log
 0.00235 x . What is the intensity, to the nearest tenth, at a depth of 40 feet?
12
(27)
Tiki went to a rock concert where the decibel level was 88. The decibel is defined by the formula
i
D  10 log , where D is the decibel level of sound, i is the intensity of the sound, and i0 = 10 -12 watt
i0
per square meter is a standardized sound level. Use this information and formula to find the intensity of
the sound at the concert.
(28)
How many years, to the nearest year, will it take the world population to double if it grows continuously
at an annual rate of 2%.
18