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
Rewrite exponential equation in logarithmic form.
811/4=3
log81 3=1/4
Condense
3ln(x-5)-ln8x
ln[(x-5)3/8x]
Solve for x:
4^x = 16
log 4^x = log 16
4
4
x = 2
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
Describe domain, range, intercept, and asymptote.
f(x)=ex
D: (-∞,∞)
R: (0,∞)
y-int: (0,1)
HA: y=0
Describe domain, range, and asymptote.
f(x)=ln(x-1)
D: (1,∞)
R: (-∞,∞)
VA: x=1
Evaluate using the change-of-base formula:
log (4)
12
log 4/log 12
= 0.56
Simplify:
e^ln(5x+2)
5x+2
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
Identify domain, range, asymptotes, and y-intercept.
f(x)=-3x+2
D: (-∞,∞)
R: (-∞, 0)
HA: y=0
y-int: (0,-9)
Rewrite the expression using properties of logarithms.
3log2(1/2)
-3
Rewrite the expression in terms of ln 4 and ln 5
ln
5/64
ln 5
ln 4^3
= ln 5 - 3ln 4
Solve for x:
ln(2x-1) = 5
e^ln(2x-1) = e^5
2x-1 = e^5
2x=149.41
x=74.71
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
$500 is invested at an annual interest rate of 10% compounded biannually. Find the balance after 10 years.
Approximately $1326.65 after ten years
Expand
ln[0.5(x2(y+z)3)/(a5b7)]
lnx+1.5ln(y+z)-5lna-7lnb
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)
Solve for x
log 25 = 2
x
x^2=25
x = 5
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
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
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
Expand the following logarithmic expression
log_2(4^2* 3^4)
log_2(4^2) + log_2(3^4)
4 + 4log_2(3)
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
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