LSP 120: Quantitative Reasoning and
Technological Literacy
Section 202
Özlem Elgün
Review from previous class:
Solving Exponential Equations
• Solving for the rate (percent change)
• Solving for the initial value (a.k.a. old value, reference
value)
• Solving for time (using Excel)
• TODAY: Solving for time (using log function)
Exponential function equation
• As with linear, there is a general equation for exponential functions. The
equation for an exponential relationship is:
y = P*(1+r)x
•
•
•
•
P = initial value (value of y when x = 0),
r is the percent change (written as a decimal),
x is the input variable (usually units of time),
Y is the output variable (i.e. population)
The equation for the above example would be
y = 192 * (1- 0.5)x
Or
y = 192 * 0.5x.
• We can use this equation to find values for y if given an x value.
• Write the exponential equation for other examples.
Solving for the rate(percent change)
What if we had to solve for the rate (percent change) but the new value is not one time period but
several time periods ahead? How do we solve for the rate then?
Example: The population of Bengal tigers was 2000 in 1973. In the year 2010 there were 4700
Bengal tigers in the world. If we assume that the percent change (rate of population growth)
was constant, at what yearly rate did the population grew?
reminder our formula is  y = P*(1+r)x
In this case:
y= 4700
P= 2000
x= number of time periods from the initial value to the new value=2010-1973=37 years
Solve for r!
4700 = 2000 *(1+r)37
4700/2000 = (1+r)37
2.35 = (1+r)37
2.351/37 = (1+r)
1.023361 =1+r
1.023361 -1 = r
0.023361 = r
0.023 = r  2.3 %
The population of Bengal tigers grew 2.3 percent every year.
Solving for the initial value (a.k.a. old value, reference value)
What if we had to solve for the initial value?
Example: The population of Bengal tigers was 4700 in 2010. If we knew that the percent change (rate of
population growth) was constant, and grew at a rate of 2.3 percent every year, what was its population
in year 1990?
reminder our formula is  y = P*(1+r)x
In this case:
y= 4700
x= number of time periods from the initial value to the new value=2010-1990=20 years
r=2.3% (0.023 in decimals)
Solve for P!
4700 = P*(1+0.023)20
4700 = P*(1.023)20
4700= P*1.575842
4700/1.575842 = P
2982.533 =P
In year 1990, there were 2982 Bengal tigers in the world
Solving for time (x) using Excel
Example: The population of Bengal tigers was 4700 in 2010. If we assume that the rate of
population growth will remain constant, and will be at a rate of 2.3 percent every year, how
many years have to pass for Bengal tiger population to reach 10,000?
bengal tiger percent
year
population change
2010
4700
0.023
Solving for time (using logarithms)
• To solve for time, you can get an approximation by
using Excel. To solve an exponential equation
algebraically for time, you must use logarithms.
• There are many properties associated with
logarithms. We will focus on the following property:
log ax = x * log a
for a>0
• This property is used to solve for the variable x
(usually time), where x is the exponent.
Solving time (x) with logarithms:
Example: The population of Bengal tigers was 4700 in 2010. If we
assume that the rate of population growth will remain constant, and
will be at a rate of 2.3 percent every year, how many years have to
pass for Bengal tiger population to reach 10,000?
Start with :
Y= P * (1 + r)X.
Fill the variables that you know. To use logarithms, x (time) must be
your “unknown” quantity.
y= 10000
P= 4700
R=0.023
Solve for x!
The equation for this situation is:
10000 = 4700 * (1+0.023)X
Solving time (x) with logarithms: Continued….
We need to solve for x:
Step 1: divide both sides by 4700
10000/4700 = (1+0.023)X
2.12766 = (1+0.023)X
USE THE LOG property you learned earlier 
log ax = x * log a
for a>0
Step 2: take the log of both sides
log(2.12766) = log (1+0.023)X
Step 3: bring the x down in front
log(2.12766) = x * log (1+0.023)
Step 4: divide both sides by log (1+.023)
log(2.12766) /log(1+.023) = x to get 33.20316
Step 5: Write out your answer in words:
If we assume that the rate of population growth will remain constant, and will be at a rate
of 2.3 percent every year 33.2 years have to pass for Bengal tiger population to reach
10,000.
Application of Exponential Models:
Carbon Dating
• A radioisotope is an atom with an unstable nucleus, which is a nucleus
characterized by excess energy which is available to be imparted either to a
newly-created radiation particle within the nucleus, or else to an atomic
electron. The radioisotope, in this process, undergoes radioactive decay,
and emits a gamma ray(s) and/or subatomic particles. These particles
constitute ionizing radiation. Radioisotopes may occur naturally, but can
also be artificially produced.
• Radiocarbon dating, or carbon dating, is a radiometric dating method that
uses the naturally occurring radioisotope carbon-14 (14C) to determine the
age of carbonaceous materials up to about 58,000 to 62,000 years
• One of the most frequent uses of radiocarbon dating is to estimate the age
of organic remains from archaeological sites.
• The Dead Sea Scrolls are a collection of 972 documents, including texts from
the Hebrew Bible, discovered between 1946 and 1956 in eleven caves in and
around the ruins of the ancient settlement of Khirbet Qumran on the
northwest shore of the Dead Sea in the West Bank.
• We date the Dead Sea Scrolls which have about 78% of the normally
occurring amount of Carbon 14 in them. Carbon 14 decays at a rate of about
1.202% per 100 years. I make a table of the form.
Years (after death)
0
100
200
.
.
% Carbon remaining
100
98.798
percent change
percent change (%)
-0.01202
-1.202%
• Using excel and extending the table we find that the Dead Sea Scrolls would
date from between 2100 to 2000 years ago.
Years (after death)
0
100
200
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
% Carbon remaining
100
98.798
97.61044804
96.43717045
95.27799567
94.13275416
93.00127845
91.88340309
90.77896458
89.68780143
88.60975405
87.54466481
86.49237794
85.45273956
84.42559763
83.41080194
82.4082041
81.41765749
80.43901725
79.47214026
78.51688513
77.57311217
76.64068337
percent change
percent change (%)
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-0.01202
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
-1.202%
Use logarithms to solve for time
• When were the dead sea scrolls created?
78 = 100*(1-0.01202)X
.78=(0.98798) X
Log (.78)= x*log (0.98798)
-0.107905397= x*(-0.005251847)
-0.107905397/-0.005251847=x
20.546=x
Since x is in units of 100 years
Dead Sea Scrolls date back 2054.6 years
(Current estimates are that a 95% confidence interval for
their date is 150 BC to 5 BC)
1. Beryllium-11 is a radioactive isotope of the alkaline metal
Beryllium. Beryllium-11 decays at a rate of 4.9% every second.
a) Assuming you started with 100%, what percent of the beryllium-11 would be remaining after
10 seconds? Either copy and paste the table or show the equation used to answer the question.
y = 100*(1-0.049)10
=100*(0.951) 10
Y=60.51
seconds
0
1
2
3
4
5
6
7
8
9
10
% Beryllium 11 remainig
100
95.1
90.4401
86.0085351
81.79411688
77.78620515
73.9746811
70.34992173
66.90277556
63.62453956
60.50693712
60.61 % Beryllium-11 remains after 10 seconds.
percent change
percent change (%)
-0.049
-0.049
-0.049
-0.049
-0.049
-0.049
-0.049
-0.049
-0.049
-0.049
-4.9%
-4.9%
-4.9%
-4.9%
-4.9%
-4.9%
-4.9%
-4.9%
-4.9%
-4.9%
1. Beryllium-11 is a radioactive isotope of the alkaline metal
Beryllium. Beryllium-11 decays at a rate of 4.9% every second.
b) How long would it take for half of the beryllium-11 to decay? This time
is called the half life. (Use the "solve using logs" process to answer the
question) Show your work.
50 =100*(1-0.049) X
.50 =(0.951) X
Log (.50) = x*log (0. 951)
-0.301029996= x*(-0.021819483)
-0.301029996 /-0.021819483=x
13.796=x
It would take 13.796 seconds for Beryllium-11 to reach its half life.