5-Minute Check on Activity 5-135-Minute Check on Activity 5-135-Minute Check on Activity 5-135-Minute Check on Activity 5-13

Click the mouse button or press the Space Bar to display the answers.Click the mouse button or press the Space Bar to display the answers.

Use the properties of logarithms to simply the following expressions:

1. ln xy

2. log 10x/z

3. ln x5

4. log y7z3

5. ln (3xy5/z4)

6. ln a + ln b

7. log 5 – log 3

= ln x + ln y

= log 10 + log x – log z = 1 + log x – log z

= 5ln x

= 7log y – 3 log z

= ln 3 + ln x + 5 ln y – 4 ln z

= ln ab

= log 5/3

Activity 5 - 14

Prison Growth

Objectives• Solve exponential equations both graphically and

algebraically

Vocabulary• None new

Activity

Your sister is a criminal justice major at WCC. The following statistics appeared in one of her required readings relating to the inmate population of US federal prisons (population in thousands).

She asks you to help analyze the prison growth situation for a project in her criminology course.

Enter the data and plot using STATPLOT. What form does the graph look like?

Year 1975 1979 1986 1990 1994 1998 2000 2003

Population 20.1 21.5 31.8 47.8 76.2 95.5 112.3 158.0

Exponential growth

Activity cont

Use an exponential regression model on your calculator to estimate the total inmate population equation.

Graph the STATPLOT and the Y1 = on the same graph

Use the model to estimate the prison population in 2010

Year 1975 1979 1986 1990 1994 1998 2000 2003

Population 20.1 21.5 31.8 47.8 76.2 95.5 112.3 158.0

P = abt = 11.36 (1.080)t

P = 11.36 (1.080)t = 11.36(1.080)40 = 246.8 (thousand)

Algebraic Method

Solving exponential problems in form abx = c1. Isolate exponential factor on one side of equation

bx = c/a

2. Take the log (or ln) of both sides of equationln bx = ln (c/a)

3. Apply log property: log bx = x log b to remove variable as an exponentx ln b = ln c – ln a

4. Solve the resulting equation for the variablex = (ln c – ln a) / ln b

Activity cont

Solve 11.36(1.080)t = 180 algebraically

180 = 11.36 (1.080)t

15.8451 = (1.080)t

ln 15.8451 = ln (1.080)t

ln 15.8451 = t ln (1.080)

ln 15.8451 ln (1.080) = 35.899 years = t

1970 + 35.899 ≈ 2006

Activity cont

Solve 11.36(1.080)t = 180 graphically

y1 = 11.36 (1.080)t

y2 = 180

Graph and find the intersection (2nd TRACE)

x = 35.899 or about 2006

Radioactive Decay

Radioactive substances, such as uranium-235, stontium-90, iodine-131, and carbon-14, decay continuously with time. If P0 represents the original amount of a radioactive substance, then the amount P present after a time t (usually measured in years) is modeled by

P = P0ekt

where k represents the rate of continuous decay

Radioactive Decay Example 1

One type of uranium decays at a rate of 0.35% per day. If 40 pounds of this uranium was found today, how much will be left after 90 days?

P = P0ekt

P = 40e-0.0035t

P = 40e-0.0035(90)

P = 40e-.315 = 29.2 lbs

Radioactive Decay Example 2

Strontium-90 decays at a rate of 2.4% per year. If 10 grams of it were initially present, how much will be left after 20 years?

P = P0ekt

P = 10e-0.024t

P = 10e-0.024(20)

P = 10e-.48 = 6.2 grams

Radioactive Decay Example 3

Strontium-90 decays at a rate of 2.4% per year. If 10 grams of it were initially present, how long until half of it is gone? (This is called half-life, a very important term with radioactive material)

P = P0ekt

5 = 10e-0.024t

0.5 = e-0.024t

ln 0.5 = -0.024t

(-0.69315)/(-0.024) = 28.88 years = t

For Note:

(ln 0.5)------------------ = half-life (decay rate)

Summary and Homework

• Summary– Solving exponential problems in form abx = c

1. Isolate exponential factor on one side of equation

2. Take the log (or ln) of both sides of equation

3. Apply log property: log bx = x log b to remove variable as an exponent

4. Solve the resulting equation for the variable

– Graph as two functions and find the intersection1. y1 = abx

2. y2 = c

• Homework– pg 669 – 72; problems 1, 4-10