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REVIEW 11.1 – 11.4. Exponential & Logarithmic Functions. Properties of Exponents. OR. Evaluate:. 1.)2.)3.) 4.)5.)6.) 7.)8.)9.) 10.)11.)12.). Express using rational exponents:. 1.)2.)3.) 4.)5.)6.) 7.)8.)9.) 10.)11.)12.). - PowerPoint PPT Presentation
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Exponential & Logarithmic Functions
REVIEW11.1 – 11.4
Properties of Exponents
€
am • an = am+ n
€
am
an = am −n
€
(am )n = am• n
€
(a • b)m = am • bm
€
ab ⎛ ⎝ ⎜
⎞ ⎠ ⎟m
=am
bm
€
a0 =1
€
a−n =1an
€
1a−n = an
€
ab ⎛ ⎝ ⎜
⎞ ⎠ ⎟−n
=ba ⎛ ⎝ ⎜
⎞ ⎠ ⎟n
€
am
n = amn
OR
Evaluate:1.) 2.) 3.)
4.) 5.) 6.)
7.) 8.) 9.)
10.) 11.) 12.)
€
1253
€
1624
€
1691
2( )0
€
2163( )2
€
811
2 − 81−1
2
€
3433( )−2
€
815
4
€
31
2 • 211
2
€
6251
4
€
2163
€
2−5 • 27
€
833
€
5
€
4
€
1
€
36
€
809
€
889
€
243
€
3 7
€
5
€
6
€
4
€
8
Express using rational exponents:1.) 2.) 3.)
4.) 5.) 6.)
7.) 8.) 9.)
10.) 11.) 12.)
€
a6b3
€
25a4b10
€
64s9t153
€
169x 5
€
15x 3y155
€
27x 4 y 34
€
x 5y 6
€
c 73
€
144x 6y10
€
27x10y 55
€
1024a3
€
36a8b54
€
a3b3
2
€
5a2b5
€
4s3t 5
€
13x5
2
€
151
5 x3
5 y 3
€
271
4 xy3
4
€
x5
2 y 3
€
c7
3
€
12x 3y 5
€
271
5 x 2y
€
32a3
2
€
61
2 a2b5
4
Express using radicals:1.) 2.) 3.)
4.) 5.) 6.)
7.) 8.) 9.)
10.) 11.) 12.)
€
x2
3
€
x4
7 y3
7
€
a1
6b2
3c1
2
€
15x1
3 y1
5
€
(5a)2
3 b5
3
€
3433( )−2
€
641
3
€
y3
2
€
21
2 a3
2b5
2
€
x2
5 y3
5
€
s2t( )1
3v2
3
€
x 6y 3( )1
2 z3
2
€
x 23
€
x 4 y 37
€
ab4c 36
€
15 x 5y 315
€
b 25a2b23
€
149
€
4
€
y 3
€
ab2 2ab
€
x 2y 35
€
s2tv 23
€
x 3 y 3z3
Simplify:1.) 2.) 3.)
4.) 5.) 6.)
7.) 8.) 9.)
10.) 11.) 12.)
€
4x 2(4 x)−2
€
(y−2)4 • y 8
€
2x( )4( )
−2
€
a3b2 • a4b5
€
(5ac)1
3 (a2c 3)1
3
€
a4b83
€
x −2( )3
• x 3( )−2
€
3y 3( ) 3y( )3
€
x 7 • x 5 • x−7 • x−5
€
6a1
3( )3
€
3x 2(3x)−2
€
2a( )1
3 a2b( )1
3
€
14
€
1
€
1256x 8
€
a3b3 ab
€
ac 5c3
€
ab2 ab23
€
1x12
€
81y 6
€
1
€
216a
€
13
€
a 2b3
General Form of Exponential Function y = b x where b > 1 Domain: All
reals Range: y > 0 x-intercept:
None y-intercept:
(0, 1)
General Form of Exponential Function y = b (x + h) + k where b > 1
h moves graph left or right (opposite way)
k move graph up or down (expected way)
So y=3(x+2) + 3 moves the graph 2 units to the left and 3 units up
(0, 1) to (– 2, 4)
Graph:
€
y =13 ⎛ ⎝ ⎜
⎞ ⎠ ⎟x
X Y
-2
-1
0
1
2
3
9
3
1
1/3
0.111
0.037
Decreasing for all of x
Graph:
€
y = 2x −1
X Y
-2
-1
0
1
2
3
0.125
0.25
0.5
1
2
4
Increasing for all of x
Graph:
€
y = −2x +1
X Y
-2
-1
0
1
2
3
-0.5
-1
-2
-4
-8
-16
Decreasing for all of x
Graph:
€
y = 2−x +1
X Y
-2
-1
0
1
2
3
8
4
2
1
0.5
0.25
Decreasing for all of x
Graph:
€
y > 2x
X Y
-2
-1
0
1
2
3
0.25
0.5
1
2
4
8
Graph:
€
y ≥ (0.5)x
X Y
-2
-1
0
1
2
3
4
2
1
0.5
0.25
0.125
Converting between Exponents & LogarithmsWRITE EACH EQUATION INTO A LOG FUNCTION:
WRITE EACH EQUATION INTO A EXPONENTIAL FUNCTION:
€
25 = 32
€
5−3 =1
125
€
6−3 =1
216
€
log2 32 = 5
€
log51
125= −3
€
log61
216= −3
€
log3 27 = 3
€
log4 16 = 2
€
log101
100= −2
€
33 = 27
€
42 =16
€
10−2 =1
100
€
log7 73
€
log10 0.001
€
3log3 6
Evaluate:
Evaluate:
Evaluate:
€
logb b−4
€
loga a
€
34 log3 4
Evaluate:
Evaluate:
Evaluate:
3
-3
6
-4
1
256
Properties of Logarithms (Shortcuts)
logb1 = 0 (because b0 = 1) logbb = 1 (because b1 = b) logb(br) = r (because br = br)
blog b M = M (because logbM = logbM)
Properties of Logarithms logb(MN)= logbM + logbN Ex: log4(15)= log45 + log43
logb(M/N)= logbM – logbN Ex: log3(50/2)= log350 – log32
logbMr = r logbM Ex: log7 (103) = 3 log7 10
If logbM = logbN Then M = N log11 (1/8) = log11 8-1
TRY THESE TO SOLVE THESE:
€
log6 x = 2
€
log5125
= y
€
logx 64 = 3
€
log4 0.25 = x
€
log4 (2x −1) = log4 16
€
log10 10 = x€
62 = xx = 36
€
5y =125
5y =152
5y = 5−2
y = −2
€
x 3 = 64x = 4
€
4 x = 0.25
4 x =14
4 x = 4−1
x = −1
€
2x −1 =162x =17
x =172
€
10x = 10
10x =101
2
x =12
First I would change to exponential form!!
€
log3(4 x −10) = log3(x −1)
€
log8(x 2 + 5x +14) = log8 0
It appears that we have 2 solutions here.If we take a closer look at the definition of a logarithm however, we will see that not only must we use positive bases, but also we see that the arguments must be positive as well. Therefore -2 is not a solution.
Our final concern then is to determine why logarithms like the one below are undefined.
Can anyone give us an explanation ?2log ( 8)
One easy explanation is to simply rewrite this logarithm in exponential form. We’ll then see why a negative value is not permitted.
2log ( 8) y
€
2y = −8
What power of 2 would gives us -8 ?
Hence expressions of this type are undefined.
SOLVE EACH LOGARITHM EQUATION:
1.) 2.)
3.) 4.)
€
4 log8 x = log8 81
€
x = ±3 €
log5(y −12) + log5(y +12) = 2
€
y =13
€
log5 42 − log5 6 = log5 k
€
k = 7 €
log10 y =14
log1016 +12
log10 49
€
y =14
€
y =13 or −13
Same Base
Solve: 4x-2 = 64x
4x-2 = (43)x
4x-2 = 43x
x–2 = 3x -2 = 2x -1 = x
If bM = bN, then M = N64 = 43
If the bases are already =, just solve the exponents
You Do
Solve 27x+3 = 9x-1
x 3 x 13 2
3x 9 2x 2
3 33 33x 9 2x 2x 9 2x 11
Review – Change Logs to Exponents
log3x = 2 logx16 = 2 log 1000 = x
32 = x, x = 9x2 = 16, x = 4
10x = 1000, x = 3
Example
7xlog25 = 3xlog25 + ½ log225 log257x = log253x + log225 ½
log257x = log253x + log251
7x = 3x + 1 4x = 11
4x
Example 1 – Solving Simple Equations
a. 2x = 32 2x = 25 x = 5 b. ln x – ln 3 = 0 ln x = ln 3 x = 3 c. = 9 3–
x = 32 x = –2
d. ex = 7 ln ex = ln 7 x = ln 7 e. ln x = –3 eln x = e–
3 x = e–3
f. log x = –1 10log x = 10–1 x = 10–1 =
g. log3 x = 4 3log3 x = 34 x = 81
Example 6 – Solving Logarithmic Equations
a. ln x = 2 eln x = e2
x = e2
b. log3(5x – 1) = log3(x + 7)
5x – 1 = x + 7
4x = 8 x = 2
Example 6 – Solving Logarithmic Equations c. log6(3x + 14) – log6 5 = log6 2x
3x + 14 = 10x
–7x = –14
x = 2
cont’d
Given the original principal, the annual interest rate, and the amount of time for each investment, and the type of compounded interest, find the amount at the end of the investment.
1.) P = $1,250; r = 8.5%; t = 3 years; quarterly
2.) P = $2,575; r = 6.25%; t = 5 years, 3 months; continuously
€
A = P 1+rn
⎛ ⎝ ⎜
⎞ ⎠ ⎟nt
€
A =1250 1+0.085
4 ⎛ ⎝ ⎜
⎞ ⎠ ⎟4(3)
€
A = $1,608.77
€
A = Pert
€
A = 2575e0.0625(5.25)
€
A = $3,575.03