Algebra 3 Assignment Sheet Logs

Alg 3 Functions 1
Algebra 3 Assignment Sheet
Logs
(1) Assignment # 1 – Exponential Equations
(2) Assignment # 2 – Logarithms
(3) Assignment # 3 – Laws of Logarithms
(4) Assignment # 4 – Calculator Problems
(5) Assignment # 5 – Review Worksheet
(6) TEST

Alg 3 Functions 2

Alg 3 Functions 3
5.1 Exponents
EXPONENTIAL EQUATIONS
I Review
x
y
b b =
x
b
y
b
b
x
y
0
b
b
x
x
bc
x
3 3
b b
b
3
2
b
b
3
2
II
x
y
If b b , then x y
x
x
If b a , then b a
If x = y, then
b
x
b
y
Ex.
x 2 5
3 3 Ex.
x+2 = 5 x = 4
x = 3
2
x
2
4
3 3
2
x 1 x 16
1) 2 32 2) 2 2
2x 1
1
3) 8 64 4) 64
x
2

Alg 3 Functions 4
1
5) 27 6) 8 4
9
2 2
x 1 x x x 5
x x x 1
x x
7) 3 3 3 8) 5 4 4 96
729
x x x x 1
9) 3 3 3 3 54
10)
2
-1
-3 -2
2 3
11)
8
1
4
2x + 1 x 3
= 16
12)
3 x
3x 2
4 9 4 + 8 = 0

Alg 3 Functions 5
III
Sketching
x
y 2 Notes
-3
-2
-1
0
1
2
3
x
y
y 2
x
-3
x
y
-2
-1
0
1
2
3
How could you sketch its inverse?

Alg 3 Functions 6
Graph
x
y = 1
Graph
y = 2
x
Graph its inverse on the same graph
* Remember how to recognize quadratic form equations

Alg 3 Functions 7
Algebra 3 Assignment # 1
Exponential Equations
(1) Solve for x please.
(a)
x
1
x
4 8
(b) 5
x
3
1
25
(c)
2x 1 x 7
2 8
(d)
9
x
2
2x
27
x
2
1
(e)
1
4
2x
1
1
x
8 16
(f) x 1
3
16
2
4
(g)
x 2 3
27
8
2 2 x 48 0
(h) 2 x 13
2 9 0
(i) 27 x 10
27 x (j) 16 x 10 16 0
4 x
(2) Sketch a graph of each of the following on the same graph.
(a)
x
y 3
(b)
y
3
x

Alg 3 Functions 8
Algebra 3 Assignment # 1
Answers
(1) (a) 2 (b) 5
(c) 20 (d) 1 , 3
(e)
9
10 (f) 3
(g) 9
4
(h) 4
(i) 0 , 3
2
(j)
1
2
,
3
2

Alg 3 Functions 9
EXPONENTIAL EQUATIONS EXTRA
(1) Evaluate each of the following numbers please.
(a)
3 1
2
9 49
2
(c)
2 2 2
2 4
(b)
2 2
2 + 2
2
1
(d)
2
1
3 2
2 3
(2) Solve each of the following equations please.
(a)
2
3
x = 25
(e)
3
2 5
x + 7 = 8
(b)
3
5 1
x = 8
(f)
3
2 4
x 6x + 9 = 27
(c)
2x + 3 x + 4
8 16 (g)
2x
8 6 8 + 8 = 0
x
(d)
1
4
2x + 1 x 3
8 = 16 (h)
3 x
3x 2
4 9 4 + 8 = 0
(3) Sketch a graph of each of the following on the same set of axes.
x
1
y = 2
and
x =
1
2
y

Alg 3 Functions 10
EXPONENTIAL EQUATIONS EXTRA
Answers
(1) (a) 27
7
(c) 1
(b) 17 2
(d) 6
(2) (a) 125 (e) 5
(b) 32 (f) 12 , –6
(c) 7 2
(g) 1 ,
2
3 3
(d) 13 8
(h) 0 , 1

Alg 3 Functions 11
5.2 LOGS
I
Def:
y
x
3 inverse? We call this exponent a logarithm
( “log” for short)
For x 0, and b 0, and b 1, the logarithmic function with base b is denoted
f(x) = log x, where y = log x if and only if x = b
b
b
y
Logarithms are exponents and follow exponent rules.
LOGARITHM FORM
EXPONENTIAL FORM
number
log n
b
p
power
p
b
power
n
number
base
base
Log form
log 8 = 3
2
log 16 = 2
4
log 1 = 0
2
1
log =
3
27
3
Exponential form
Exponential form
3
3 27
2 1
5
25
3
2 8
Log form
x
4 1

Alg 3 Functions 12
RULES
x
1) n > 0, p must be real 4) log b = x
b
2) b 1 5) log 1 = 0
b
b 1 b 2
1 2
log
bx
3) If log x = log x , then x = x 6) b = x
EXAMPLES
1. log 25 =
5
2. log 3 =
3
3. log 16 =
b
2
4. log b =
b
5. log x = 1
6
6. Find the inverse of log 2x + 4
2

Alg 3 Functions 13
MORE EXAMPLES
log 81 x log 1 c
3 8
log 4 y log 9 d
2 27
log5
25 z log9 3
log x
1
2
log 8 p log x
8 125
2
3
log 6 a
6

Alg 3 Functions 14
Examples
1)
8
log =
27
2
3
1 3
2) log =
b
27 2
1
log (log ) x 4) log.00110
81
3)
3 1
4 27
y

Alg 3 Functions 15
Compound interest arises when interest is added to the principal, so that from that moment on, the
interest that has been added also itself earns interest. This addition of interest to the principal is called
compounding. A bank account, for example, may have its interest compounded every year: in this case,
an account with $1000 initial principal and 20% interest per year would have a balance of $1200 at the
end of the first year, $1440 at the end of the second year, and so on.
20% growth on a $1000 investment
P = principal amount (the initial amount you borrow or deposit)
r = annual rate of interest (as a decimal)
t = number of years the amount is deposited or borrowed for.
A = amount of money accumulated after n years, including interest.
n = number of times the interest is compounded per year
Wikipedia

Alg 3 Functions 16
http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_Exponents_e.xml
Natural Log of (e) = ln (e)
A. A valuable number for calculus and compound interest
1
1 + n
n
If you took this to many places on the calculator you would see it approaches the number 2.71828…
This number is designated as “e”.
It is used often with logs and has its own designation.
x
If e is the base , then the exponential function is y = e , and
the log function is expressed as y = ln e . It is assumed the base is e when using ln instead of log.
B. Properties of “e”
C. Examples
1) 4 ln( e) = x 2)
ln e 3
2 = x

Alg 3 Functions 17
Algebra 3 Assignment # 2
Definition of the logarithm
Solve for x please.
(1) log 4
64 x
*(10)
3 2
5ln e
= x
log (11) 6
log 2
(2) x 2
x
1
25
log (12) x
log 36
216 x
(3) 9 2
(4) log 3
x 2
(13)
log
4 x
3
2
log (14) 25
log 8
4 2 x
(5) 125 x
(6)
2
log
8 x
(15)
3
log
x 6
1
2
(7) log 27
81 x
*(16)
2
ln e
x = 5
(8) log 7
7 x
*(17)
log
4
log
2
x
1
2
(9)
3
log
16 x
*(18)
4
log
16
log
x
9
1
4

Alg 3 Functions 18
Algebra 3 Assignment # 2
Answers
(1) 3 (10) 10 3
(2) 36 (11) 5
(3) 3 (12) 2
3
(4) 9
1
(13) 8
1
(5) 2
3
(14) 6
5
(6) 4 (15) 36
1
(7) 3
4
(16) 25
(8) 2
1
(17) 4
(9) 8
1
(18) 3

Alg 3 Functions 19
5.3 LAWS OF LOGS
Properties of Logs
I LAWS
log mn log m log n
b b b
m
logb logbm logbn
n
log m
b
p
plog m
b
II Equations
If log m log n then m n
b
log =log
If log m n then
b
(log number)
If a = b, then x = x
a
b
b
b
n
m
1) log 6 log 2 log 3
3 3 3
5)
2
log (2n + 20) = log (32 5n )
5 5
log
x
5
log x
log 5
2)
5 5 5
l g 4 3log 4
o 6)
3
3)
4 4
3 3
x = 4 , then x = 4
Now go backwards
III Examples
7) log6 48 log6 w log6
4 4) log2 3 log2 7 log2
x

Alg 3 Functions 20
1
1 1
8) log10 m log10
81
9) log7 m log7 64 log7121
2
3 2
10) log 2( y 2) 1 log 2 y 2
11)
2
log 4 = x
8
12) log10 y log10
y 21 2
Uh, oh, what is this????
13)
log3
5
log169
3 = x 14) 4
15)
ln 3
e = 16)
2
ln e =

Alg 3 Functions 21
Change of Base Rule:
What happens if you have a problem where the bases are not the same?
Example: log
23 = x

Alg 3 Functions 22
Algebra 3 Assignment # 7
Properties of Logarithms
(1) Evaluate each of the following please.
log
(a) 7 6
log
7 (b) 25 36
5
log
(c) 8 27 log4
25
4 + 8 (d)
2ln 8 3ln 4
e
(2) Solve for x please.
(a)
log 2x 1 3x 6
3
log (b)
3
log 2
10 x + 9x = 1
(c)
log x = 4 3
5
log (d)
5
log x = log 144 log 8
1 1
9 2 9 3 9
(e)
log 7 +
3
log x 2 = 6x
3
log (f) ln 15 + ln 14 ln 105 = ln x
3
(g)
log x 1 +
10
log x + 2 = 7
10
log (h)
7
log x + 3 +
3
log x 3 = 16
3
log
3
(i)
log x + 1
8
log x = 6x + 2
8
log (j)
8
log x + 3 +
3
log 4x 1 = 12
3
log
3
(k)
2
x x 2x 5 =
2
8 8 3
x log 9 log 4
log log (l) 125 = 8
4
3
9

Alg 3 Functions 23
Algebra 3 Assignment # 7
Answers
(1) (a) 6 (b) 6
(c) 134 (d) 1
(2) (a) 7 (b) 10 , 1
(c) 9 (d) 6
(e) 14 (f) 2
(g) 3 (h) 5
(i) 1 3
(j) 1
(k) 4 , 5 (l) 2 3

Alg 3 Functions 24
MORE LOG EQUATIONS
NOTE:
1
4
2
2 , 2
2
4 ,
1
1
2 2
3
9 3 , 3 9 , 3 27
I Solve for x please
1)
log 27 x 2)
4
9
2
1 1
3 3
log 12x log 20x
9 1
3) log1
x 2
4)
3
2
4 4
2 log x 5log x 12 0
log 6
log9
4
2
5) 9 x 6)
4 b
log4
x
7) 8 8 8)
log 5
2x 2
125 16

Alg 3 Functions 25
Algebra 3 Assignment # 8
Properties of Logarithms 2
(1)
Solve for x please.
2
x 1 5x 11 =
1
4 4 2
log log
(2)
2
6 6 6
log log log
3x 5 x 1 = x 1
(3)
2
ln 4x + 1 + ln x + x = ln 19x 9
(4)
2 2
3x 7 x x 1 =
2
8 8 3
log log
(5)
2
ln x + 4 + ln 3x 4 = ln 17x 18
2ln 3 log9
25
(6) log x + 1 + x 5 = e 3
2
log
2
* (7)
log x 5 = x + 7
3
log ( change of base)
9
(8)
2
3 log8 x log
8
x 2 = 0
(9)
2
4 4
2 log x + 5log x = 0
(10)
ln
2
2 x = 3

Alg 3 Functions 26
Algebra 3 Assignment # 8
Answers
(1) 3 , 7
(2) 2 , 3 , reject 5
(3)
3
1 , , reject 3
4
(4) 3 , (reject 1)
(5)
2 , reject 1 ,
1
3
(6) 7 , (reject 3)
(7) 9 , (reject 2)
(8)
8 ,
1
4
(9) 1
32 , 1
(10) Ø

Alg 3 Functions 27
Using calculators
1)
x
3 8 2) log2
6
x
3)
3y y 3
6 8 4)
4x 7 2x 3
3 4
5) 1) If log 3 = .4771 and log 2 = .3010
b b
find a) log 12 b) log 1.5
b
b

Alg 3 Functions 28
6)
x
.03 5

Alg 3 Functions 29
Algebra 3 Assignment # 9
(1) Use a calculator to solve each of the following correct to 4 decimal places please.
(a)
x
5 = 20 (b)
3x + 1 1 x
4 = 9
(c)
log 3 18 = x
(d) log 7 x = 1.432
log
ln (f) 5
3 x = 11
(e) x = 1.432
(g)
x
0.3 > 7 (h)
2
2 x 5 x 3 = 0
ln ln
(2) Let
log 10 2 = p and
log 10 3 = q . Evaluate each of the following in terms of p and q.
(a)
(c)
(e)
log 10 6
(b)
3 3
log (d)
10 5
16
log 10 0.5
(f)
log 10 72
log 10 90
log 10 5
(3) Simplify the following expression please.
log 125
4
log 32 7
49
log
25
(4) The magnitude of an earthquake is measured using the Richter scale;
2 E
M = log ,
3 4.4
10
Where M is the magnitude of the earthquake, and E is the seismic energy released by the
15
1.12 x 10
earthquake (in joules). The 1989 San Francisco earthquake released approximately
joules. Calculate the magnitude of the earthquake using the Richter scale. How much energy
would be released (in joules) by an earthquake which measures 8.3 on the Richter scale?

Alg 3 Functions 30
Algebra 3 Assignment # 9
Answers
(1) (a) 1.8614 (b) 0.1276
(c) 2.6309 (d) 16.2248
(e) 4.1871 (f) 5.1388
(g) x < 1.6162 (h) 0.6065 , 20.0855
(2) (a) p + q (b) 3p + 2q
(c) 3 q 4 p (d) 2q + 1
2 5
(e) p (f) 1 p
(3) 15 8
(4) 7.1 ,
16
7.079 x 10 joules

Alg 3 Functions 31
Algebra 3 Assignment # 10 ─ Review Worksheet
(1) Solve for x please.
(a)
2–x 2x+1
log 4
9 = 27 (i) 9
9
= x
(b)
2x–5 x+1
log 4 log 4
8 = 16 (j) 3
9
– 9
3
= x
(c)
2x
1–x
4 8 = 1
(k)
16
2
3 log 8(x) – 2 log 8(x) – 1 = 0
(d)
log
3
8 4 = x (l) 5 5
log 2x 3 = log 1– x
1
(e) log 1 x = – (m) log
4 2
2 x 1 + log2
3x–1 = 5
(f)
4
log 16 = – (n) log2 x – 3 – log2 x+1 = log2
8
3
x
(g)
log x .125 = 3 (o) log7 x 1 + log7 x + log7 2x 1 = log7
30
(h) log3 log8
x = –1 (p)
log 2 log 6 log x
4 2
4
4 + 4 = 8
(2) Use a calculator to solve for x. Express answers correct to 3 decimal places.
(a)
x
3 = 8 (b)
3x–2 1–x
2 = 5
(c) log3
2 = x (d)
3
ln x 4ln
x = 0

Alg 3 Functions 32
Algebra 3 Assignment # 10 ─ Review Worksheet
Answers
(1) (a)
8
1
(i) 4
(b)
19
2
(j) –14
(c) –3 (k) 8 , 2
1
(d)
9
2
(l)
2
–
3
(e) 2 (m) 3
(f)
8
1
(n)
(g) 2
1
(o) 2
(h) 2 (p) 4
(2) (a) 1.893 (b) 0.812
(c) 0.631 (d) 1 , 7.389 , 0.135

Alg 3 Functions 33
Logs & Twigs
(1)
log x + 3 +
3
log 4x 1 = 12
3
log
3
1 , reject
15
4
(2)
2
x x 2x 5 =
2
8 8 3
log log 4 , 5
(3)
2
ln 4x + 1 + ln x + x = ln 19x 9
3
1 , , reject 3
4
(4)
2
6 6 6
log 3x 5 log x 1 = log x 1 2 , 3 , reject 5
(5)
2
x 1 5x 11 =
1
4 4 2
log log 3 , 7
(6)
2 2
3x 7 x x 1 =
2
8 8 3
log log 3 , (reject 1)
(7)
2
ln x + 4 + ln 3x + 4 = ln 17x 18
2 , reject 1 ,
1
3
2ln 3 log9
25
(8) x + 1 + x 5 = e 3
2 2
log log 7 , (reject 3)
(9)
log x 5 = x + 7
3
log 9 , (reject 2)
9