Chapter 6 Review

Transformation

100

How is the graph changing?

h(x) = 3^x + 5

Shifts up by 5 units

200

How is the graph changing?

g(x) = 2^x - 1

Shifts down by 1 unit

300

How is the graph changing?

f(x) = -4^x+9

The function flips and then shifts left by 9 units

400

How is the graph changing?

h(x) = 5^x-1 + 8

Shifts right 1 unit and up 8 units

500

How is the graph changing?

g(x) = 2(3^x) - 4

The graph stretches and shifts down 4 units

Exponential Growth & Decay

100

Determine whether the function represents exponential growth or decay.

y = 5(3.7)^x

Growth

200

Determine whether the function represents exponential growth or decay.

y = 287(0.75)^x

Decay

300

A car was purchase for $34,000. The function of y=34(0.9)^x can be used to model the value of the car (in thousands of dollars) x years after it was purchased. Does this function represent growth or decay?

Decay

400

A car was purchase for $28,000. The function of y=28(0.7)^x can be used to model the value of the car (in thousands of dollars) x years after it was purchased. What is the rate of growth/decay?

30% decrease each year (decay)

500

Three-hundred bunnies were released on January 1st, 2018. The function f(x)=300(1.08)^x can be used to model the number of bunnies in the region after x years after 2018. What is the rate of growth/decay?

8% increase each year (growth)

Compound Interest

100

$50,000 is invested in an account that earns 4.63% annual interest, compounded monthly. What is the equation used to calculate the final amount after t time periods?

A = P(1+(r/n))^nt

200

$100,000 is invested in an account that earns 3.9% annual interest, compounded daily. What is the n?

Daily; 365

300

$25,000 is invested in an account that earns 2.4% annual interest, compounded semi-annually. Set up the equation for the value of the account after 9 years.

A = 25,000(1+(.024/2))²⁽⁹⁾

400

$17,200 is invested in an account that earns 3.3% annual interest, compounded continuously. What is the value of the account after 11 years?

$24,727.34

500

$110,000 is invested in an account that earns 4.1% annual interest, compounded continuously. What is the value of the account after 3 years?

$124,397.29

Log Properties

100

Expand the expression:

log₃(x⁴)

4log₃(x)

200

Expand the expression

ln(xy²)

ln(x)+2ln(y)

300

Write as a single logarithm:

log₉(r)-3log₉(s)-2log₉(t)

log₉(r/st)

400

Solve for x:

log₃(3)^x = x² - 42

x = 7 and x = -6

500

Solve for x:

ln(x² - 52) = ln(9)

x = 13 and x = -4

Different Bases & Approximations

100

Given the following:    log₃5 = 1.465    log₃6 = 1.631

log₂(25)

approximately 2.93

200

Given the following:   log₂7 = 2.8074    log₂9 = 3.1699

log₂(63)

approximately 5.9773

300

Solve for x:

log₂(6x+8)=5

x = 4

400

Solve for x:

5^x+1 = 3^x

approximately -3.1507

500

Solve for x:

2^2x-5 = 3^x

approximately 12.0471