Topic 4: Indices and Logarithms Lecture Notes: section …. Solve for x x = log( .) log( ) 1 1 100...
Transcript of Topic 4: Indices and Logarithms Lecture Notes: section …. Solve for x x = log( .) log( ) 1 1 100...
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Topic 4: Indices and Logarithms Lecture Notes: section 3.1 Indices section 3.2 Logarithms Jacques Text Book (edition 4): section 2.3 & 2.4 Indices & Logarithms
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INDICES Any expression written as an is defined as the variable a raised to the power of the number n n is called a power, an index or an exponent of a e.g. where n is a positive whole number, a1 = a a2 = a × a a3 = a × a × a an = a × a × a × a……n times
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Indices satisfy the following rules: 1) where n is positive whole number an = a × a × a × a……n times e.g. 23 = 2 × 2 × 2 = 8 2) Negative powers…..
a-n = na1
e.g. a-2 = 2
1a
e.g. where a = 2
2-1 = 21
or 2-2 = 41
221
=×
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3) A Zero power a0 = 1 e.g. 80 = 1 4) A Fractional power
n aa n =1
e.g. 3999 221
===
288 331
==
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All indices satisfy the following rules in mathematical applications Rule 1
am. an = am+n
e.g. 22 . 23 = 25 = 32
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Rule 2
na
ma = am - n
e.g. 2
3
22
= 23-2 = 21 = 2 ________________________________ note: if m = n,
then nama
= am – n = a0 = 1 ________________________________
note: na
ma− = am – (-n) = am+n
________________________________
note: na
ma −
= a-m – n = nma +1
_________________________________
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Rule 3 (am)n = am.n
e.g. (23)2 = 26 = 64
Rule 4 an. bn = (ab)n
e.g. 32 × 42 = (3×4)2 = 122 = 144 Likewise,
n
ba
nb
na⎟⎠⎞
⎜⎝⎛= if b≠0
e.g.
4236
36 2
2
2
2==⎟
⎠⎞
⎜⎝⎛=
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Simplify the following using the above Rules: 1) b = x1/4 × x3/4
2) b = x2 ÷ x3/2
3) b = (x3/4)8
4) b = yxyx
4
32
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LOGARITHMS A Logarithm is a mirror image of an index If m = bn then logbm = n The log of m to base b is n If y = xn then n = logx y The log of y to the base x is n e.g. 1000 = 103 then 3 = log10 1000 0.01 = 10-2 then –2 = log10 0.01
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Evaluate the following: 1) x = log39 the log of m to base b = n then m = bn
the log of 9 to base 3 = x then
9 = 3x 9 = 3 × 3 = 32 x = 2
2) x = log42 the log of m to base b = n then m = bn
the log of 2 to base 4 = x then
2 = 4x 2 = √4 = 41/2 x = 1/2
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Using Rules of Indices, the following rules of logs apply 1) logb(x × y) = logb x + logb y
eg. ( ) 3232 101010 logloglog +=× 2) logb ⎟⎟
⎠
⎞⎜⎜⎝
⎛yx = logb x – logb y
eg. 2323
101010 logloglog −=⎟⎠⎞
⎜⎝⎛
3) logb xm = m. logb x
e.g. 323 102
10 loglog =
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From the aboverules, it follows that (1) logb 1 = 0
(since => 1 = bx, hence x must=0)
e.g. log101=0 and therefore,
logb ( )x1
= - logb x e.g. log10 (1/3) = - log103 (2) logb b = 1
(since => b = bx, hence x must = 1) e.g. log10 10 = 1
(3) logb ( )n x = n1
logb x
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A Note of Caution:
• All logs must be to the same base in applying the rules and solving for values
• The most common base for logarithms are logs to the base 10, or logs to the base e (e = 2.718281…)
• Logs to the base e are called Natural Logarithms
logex = ln x If y = exp(x) = ex Then loge y = x or ln y = x
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Features of y = ex • non-linear • always positive • as ↑ x get ↑ y and ↑ slope of graph (gets
steeper)
0.00
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
0 0.02 0.05 0.1 0.15 0.2 0.25 0.5 0.75 1 1.25 1.5 1.75 2
x
y=ex
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Logs can be used to solve algebraic equations where the unknown variable appears as a power
An Example : Find the value of x
200(1.1)x = 20000
Simplify
divide across by 200
(1.1)x = 100
1. to find x, rewrite equation so that it is no longer a power
Take logs of both sides
log(1.1)x = log(100)
rule 3 => x.log(1.1) = log(100)
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2. Solve for x
x = ).log()log(
11100
no matter what base we evaluate the logs, providing the same base is applied both to the top and bottom of the equation
3. Find the value of x by evaluating logs using (for example) base 10
x = ).log()log(
11100
= 041402
. = 48.32
4. Check the solution
200(1.1)x = 20000
200(1.1)48.32 = 20004
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Another Example: Find the value of x
5x = 2(3)x
1. rewrite equation so x is not a power Take logs of both sides log(5x) = log(2×3x)
rule 1 => log 5x = log 2 + log 3x
rule 3 => x.log 5 = log 2 + x.log 3
2. Solve for x
x [log 5 – log 3] = log 2
rule 2 => x[log ⎟⎠⎞
⎜⎝⎛
35 ] = log 2
x = )log()log(
352
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3. Find the value of x by evaluating logs using (for example) base 10
x = )log()log(
352
= 22190301030..
= 1.36
4. Check the solution
5x = 2(3)x ⇒ 51.36 = 2(3)1.36 ⇒ 8.92
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An Economics Example 1 Y= f(K, L) = A KαLβ
Y*= f(λK, λL) = A (λK)α( λL)β Y*= A KαLβλα λ β = Yλα+β
α+β = 1 Constant Returns to Scale α+β > 1 Increasing Returns to Scale α+β < 1 Decreasing Returns to Scale Homogeneous of Degree r if:
f(λX, λZ ) = λr f(X, Z) = λr Y Homogenous function if by scaling all variables by λ, can write Y in terms of λr
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An Economics Example 2
National Income = £30,000 mill in 1964. It grows at 4% p.a. Y = income (units of £10,000 mill)
1964: Y = 3 1965: Y = 3(1.04) 1966: Y = 3(1.04)2 1984: Y = 3(1.04)20
Compute directly using calculator or
Express in terms of logs and solve
1984: logY = log{3×(1.04)20}
logY = log3 + log{(1.04)20}
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logY = log3 + 20.log(1.04)
evaluate to the base 10
logY = 0.47712 + 20(0.01703)
logY = 0.817788
Find the anti-log of the solution:
Y = 6.5733
In 1984, Y = £65733 mill
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Topic 3: Rules of Indices and Logs Some Practice Questions:
1. Use the rules of indices to simplify each of the following and where possible evaluate:
(i) 6
25
33.3
(ii) 2
24
56.5 −
(iii) xxx 26. −
(iv) ( )234x
(v) 2
2
xxy
(vi) 24
6
5315
xxx
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2. Solve the following equations: (i) x=64log4
(ii) x=⎟⎠⎞
⎜⎝⎛
271log3
(iii) 10ln4=x
(iv) 255 =x (v) 1004 =xe (vi) 10012 =−xe