AFM
Log Test Review
Name: ________________________________
Graph the following equations. Label a point on the graph and name the asymptote.
f( x ) = e(x – 1)
1. f( x ) = 3x
2.
point = ________ Asymptote _______________
point = ________ Asymptote ______________
3. f( x ) = 2x + 1
4.
point = ________ Asymptote _______________
point = ________ Asymptote ______________
5. f( x ) = log2(x)
6.
point = ________ Asymptote _______________
point = ________ Asymptote ______________
7. f( x ) = ln (x ) + 2
8.
point = ________ Asymptote _______________
point = ________ Asymptote ______________
f( x ) = 3(x + 1) – 1
f( x ) = log (x – 1)
f( x ) = log (x – 2) – 2
Write in exponential form
9.
log5125 = 3
10.
ln y = 5
11.
12.
log 4 = x
14.
xy = a
16.
e9 = x
Fill in the blank.
17.
log12129 = ________
18.
log1251 = ________
19.
log
20.
ln e = ________
21.
log12 _____ = 1
22.
e ln 1,341 = ________
24.
ln e
x
log7a = 5
Write in logarithmic form.
13.
52 = 25
15.
93 = 729
_________
=0
Expand the following log functions.
23.
log (2x)3
Write as a single logarithm.
25.
3 log3 ( 2) - ½ log3 4
26.
log
x 1
x 1
2
Solve the following equations.
27.
19 e (x – 3) = 38
28.
e (5 + x) = -3
29.
10x(2x) – x2(2x) = 0
30.
e2x – 5ex + 6 = 0
31.
log2 ( x – 3) = 8
32.
8 ln x = 9
33.
log (x – 3) + log ( x – 4) = 1
34.
log ( x – 3) + log 8 = log 16 – log (x)
Compound interest
Exponential Growth
n times a year A(t) = P ( 1 + r/n)nt
continuous
A = Pert
n(t) = n0ert
36.
A man invests $5,000 in an account
that pays 5% interest rate per year that is
compounded semiannually. How long will it
take for to have 12,000 in the account?
35.
A man invests $ 17, 300 for 3 years
in an account that pays 11% interest
rate per year. Find the amount he makes
if it is compounded monthly and
continuously.
37.
The number of people in a country is currently 5.2 million. If the country’s population grows exponentially at
a rate of 15%, how long will it take for the country’s population to reach 10 million?