3.1 Exponential Functions and Their Graphs
3.2 Logarithmic Functions and Their Graphs
3.3 Properties of Logarithms
3.4 Solving Exponential and Logarithmic Equations
3.5 Exponential and Logarithmic Models
100

Describe the transformations.

f(x)= 2x

g(x)= -2(x+5)-4

Reflection in x-axis

Horizontal shift 5 units left

Vertical shift 4 units down

100

Rewrite exponential equation in logarithmic form.

811/4=3

log81 3=1/4

100

Condense

3ln(x-5)-ln8x

ln[(x-5)3/8x]

100

Solve for x:

4^x = 16

log      4^x = log          16

                            4

      4

x = 2

100

Find the amount of time to double if the investment is compounded continuously. 

Initial Investment: 10,000

Annual Rate = 3.5%

Pe^rt

10,000e^0.035t

20,000 = 10,000e^0.035t

2 = e^0.035t

ln 2 = 0.035t

t = 19.8 years

200

Describe domain, range, intercept, and asymptote.

f(x)=ex

D: (-∞,∞)

R: (0,∞)

y-int: (0,1)

HA: y=0

200

Describe domain, range, and asymptote. 

f(x)=ln(x-1)

D: (1,∞)

R: (-∞,∞)

VA: x=1

200

Evaluate using the change-of-base formula:


log   (4)

   12

log 4/log 12

= 0.56

200

Simplify:

e^ln(5x+2)

5x+2

200

Population = 134.0e^kt (*in thousands*)

t=0 ; 1990

2000 population = 180,000

Find the value of K

180 = 134e^10k

1.34 = e^10k

ln 1.34 = 10k

0.29 = 10k

k=0.029

300

Identify domain, range, asymptotes, and  y-intercept.

f(x)=-3x+2

D: (-∞,∞)

R: (-∞, 0)

HA: y=0

y-int: (0,-9)

300

Rewrite the expression using properties of logarithms.

3log2(1/2)

-3

300

Rewrite the expression in terms of ln 4 and ln 5


ln

    5/64

ln 5

ln 4^3

= ln 5 - 3ln 4

300

Solve for x:

ln(2x-1) = 5

e^ln(2x-1) = e^5

2x-1 = e^5

2x=149.41

x=74.71

300

Population in Tilted Towers = 258.0e^kt

t = 0;1990

2000 population = 478,000

a) Find the value of k

b) Find the population in 2010

a) 478 = 258e^10k

    1.85 = e^10k

    ln 1.85 = 10k

    0.62 = 10k

    k = 0.062


b) 258e^0.062(20)

    258e^1.24

    258(3.46)

    891.5

    About 891,500

400

$500 is invested at an annual interest rate of 10% compounded biannually. Find the balance after 10 years. 

Approximately $1326.65 after ten years

400

Expand

ln[0.5(x2(y+z)3)/(a5b7)]

lnx+1.5ln(y+z)-5lna-7lnb

400

Prove (hint: let x=logau and y=logav and change to exponential forms)

loga(uv)=logau+logav

ax=u    ay=v

uv=axay

uv=ax+y

loga(uv)=x+y

loga(uv)=loga(u)+loga(v)

400

Solve for x


log        25 = 2

       x

x^2=25

x = 5

400

The amount of John Wicks in the real word can be modeled by: y= 663/1 + 72e^-0.547t

Using the model, find the population for the 19th hour 

663/1+72e^-0.547(19)

663/1.0022

661.54 John Wicks


500

The approximate number of fruit flies in an experimental population after t hours is given by Q(t)=20e0.03t, where t≥ 0.

a. Find initial number of flies.

b. How large is the population after 72 hours?

a. 20 flies

b. roughly 173 flies

500

Make describe the transformations and determine the domain, range, and asymptote.

 f(x)=3ln(x+6)-4

Horizontal translation 6 units left

Vertical stretch by a factor of 3

Vertical translation 4 units down

D: (-6, ∞)

R: (-∞,∞)

VA: x=-6

500

Expand the following logarithmic expression

log_2(4^2* 3^4)

log_2(4^2) + log_2(3^4)

4 + 4log_2(3)

500

Solve for x

5(2^3-x)-13 = 100

5(2^3-x)-13 = 100

5(2^3-x) = 113

2^3-x = 22.6

log          2^3-x = log           22.6

                                    2

        2

3-x = 4.5

-x = 1.5

x = -1.5

500

The amount of Defaults skins are released to the real world into a city honored for them. The carrying capacity is 1000. This logistic curve can be modeled by p(t) = 1000/1+9e^-0.1656t


After how many months will the population hit 500?

500 = 1000/1+9e^-0.1656t

500+4500e^-0.1656t = 1000

4500e^-0.1656t = 500

e^-0.1656t = 0.11

-0.1656t = -2.21

t = 13 months