Logarithms A Quick Review of Exponents Exponent 5 7 Base - - PowerPoint PPT Presentation

Logarithms

A Quick Review of Exponents

57 Exponent Base

Exponents have two parts:

• 1. Base: The number of variable being multiplied
• 2. Exponent (or power or index): number of times it is multiplied

57 = 5 ·5 ·5 ·5 ·5 ·5 ·5 = 78125 a3 = a ·a ·a

Introduction

Now let’s consider a simple exponent: 23 = 8 This exponent says that 2 (the base) raised to the power of 3 (the exponent) gives us 8. So, if I were to ask the question:

What power of 2 gives 8?

We know the answer is 3.

Intro, continued

Here’s another question:

What power of 3 gives 81?

We could write this question as 3? = 81

• r, we could use “power” notation:

power3(81) = ?

(What power of base 3 yields 81?)

The answer is 4: 34 = 81 or power3(81) = 4

• r

power334 = 4

Logarithm Definition

Here are more examples power2(16) = 4 power5(125) = 3 power1/2(4) = -2 Mathematicians do not use the word “power” in this context, they prefer the word “logarithm” and abbreviate it as “log” log2(16) = 4 log5(125) = 3 log1/2(4) = -2

A Logarithm Pattern

log2(16) = 4 = log224 log5(125) = 3 = log553 log1/2(4) = -2 = log1/2(1 /2)-2 Logarithms undo exponents! logb(ba) = a Example : log10105 = 5 (Challenge) This also works in reverse: blogb(a) = a Example : 10log10(100) = 10log10102 = 102 = 100

Why Logarithms Matter: Exponential Growth and Decay

Example: growth of bacteria in a petri dish B = 25 ·10h where B is the number of bacteria after h hours

Exponential Growth and Decay

One of the questions we could ask about this growth is: When does the number of bacteria equal 100,000? Graph says: ≈ 3.6 hours How can we get an exact answer? 100000 = 25 ·10h 10h = 4000 Use logarithms (base 10) log1010h = log10(4000) h = log10(4000) Using a calculator: h = 3.602059991 hours

Why Logarithms Matter: Sound Volume

The human threshold of hearing (ToH) is 10-12 W /m2. The threshold of hearing pain is 10 W /m2. Range is very large so a logarithmic scale is used

ToH is assigned 0 (zero) Bels A sound that is 10 times more intense is 1 Bel (or 10 decibels) A sound that is 100 times more intense is 2 Bels (20 decibels)

Sound Volume

Intensity Intensity Level

• No. of Times

Source W/m2

(dB)

Greater than ToH Threshold of Hearing (TOH)

• 1. × 10-12

100 Rustling Leaves

• 1. × 10-11

10 101 Whisper

• 1. × 10-10

20 102 Normal Conversation

• 1. × 10-6

60 106 Busy Street Traffic

• 1. × 10-5

70 107 Vacuum Cleaner

• 1. × 10-4

80 108 Large Orchestra 6.3 × 10-3 98 109.8 iPod at Maximum Level

• 1. × 10-2

100 1010 Front Rows of Rock Concert

• 1. × 10-1

110 1011 Threshold of Pain

• 1. × 101

130 1013 Military Jet Takeoff

• 1. × 102

140 1014 Instant Perforation of Eardrum

• 1. × 104

160 1016

Sound Volume Example

A mosquito’s buzz is often rated with a decibel rating of 40 dB. Normal conversation is often rated at 60 dB. How many times more intense is normal conversation compared to a mosquito’s buzz? A mosquito’s buzz is 104times greater than the ToH and the normal conversation is 106times greater, so the ratio is: 106 104 = 102 = 100 Therefore, a normal conversation is 100 times more intense than a mosquito’s buzz. Another way of getting to this answer is to note that the difference in intensity is 20 dB or 2 Bels: 102 = 100

Why Logarithms Matter: Acids, Bases and the pH Scale

Water Chemistry: Hydroxide ions, OH- and hydrogen ions, H+:

Acids donate hydrogen ions Bases donate hydroxide ions

Fact: a strongly acidic solution can have 100,000,000,000,000 times more hydrogen ions than a strongly alkaline solution. log10(100000000000000) = log101014 = 14 The pH scale goes from 0 (most acidic solution) to 14 (most alkaline solution)

Acids, Bases and the pH Scale

H+ Concentration pH Value Relative to Pure Water Example 10000000 battery acid 1 1000000 gastric acid 2 100000 lemon juice,vinegar 3 10000

• range juice,soda

4 1000 tomato juice,acid rain 5 100 black coffee,bananas 6 10 milk,saliva 7 1 pure water 8 0.1 sea water,eggs 9 0.01 baking soda 10 0.001 Great Salt Lake, milk of magnesia 11 0.000 1 ammonia solution 12 0.000 01 soapy water 13 0.000 001 bleach,oven cleaner 14 0.0000 001 liquid drain cleaner

Why Logarithms Matter: Earthquakes

The Richter Scale, used to measure the energy of an earthquake, is a logarithmic scale (base 10).

Magnitude of 5 is 10 times greater than magnitude of 4 Largest earthquake: Chile, 1960: 9.5 Recent California quake: 7.1

109.5 107.1 = 109.5-7.1 = 102.4 = 251.189

Earthquakes

If the Chilean earthquake had a magnitude of 9.5, what magnitude of earthquake is 500 times less? The question is asking us to find m: 109.5 10m = 500 If we rearrange and use logs, we can get the answer: 10m = 109.5 500 log(10m) = log 109.5 500 m = 6.801

Why Logarithms Matter: Large-Magnitude Math

Try 500! on a calculator. 500! = 500 ·499 ·498 ·497 · ⋯ ·3 ·2 ·1 Interesting log property (we’ll prove it soon): log10(500!) = log10 500 + log10 499 + log10 498 + log10 497 + ⋯ + log10 3 + log10 2 + log10 1 Can easily use a spreadsheet to get: log10(500!) = 1134.086409 If we undo the logarithm with an exponent we get: 500! = 101134.086409 = 101134+0.086409 = 101134 ·100.086409 = 1.220136826 ×101134

Common Logarithms

Logarithms to the base 10 are called common logarithms (presumably because our counting system is base 10 and it’s pretty common). So when we write a logarithm statement and there is no base specified, we automatically assume it is in base 10): log(x) = log10(x) Most calculators will have a “log” or “Log” or “LOG” button. This button will give logs in base 10.

Natural Logarithms

Exponents with a base of e occur frequently in the natural world. This means we often encounter functions

• f the form:

f (x) = a ex

• r

f (x) = a e-x where e = 2.71828182846... For example: loge(25) = 3.218875825 Since natural logarithms occur frequently in mathematical analysis, the symbol loge is given its own separate notation: ln This is pronounced as “ellen” in the US and as “lawn” in Canada.

Graphs of Logarithmic Functions

A Special Case: logb(1)

All of the plots in the above graph intersect at (1, 0). Why? To answer that question, let’s go back to our

• riginal “power” notation for logs:

powerb(1) = ? In words: what power of b gives us 1? b? = 1 The answer, of course, is 0: b0 = 1 We can write this as: logb(1) = 0

Domain and Range of logb(x)

From the graphs, it should be clear that the domain of logb(x) is D = {x : x > 0} = (0, ∞) The domain of the logb(x) function must be restricted to be greater than zero: there is no real-numbered exponent that will yield a negative number: b? = -1 In other words, powerb(x) = logb(x) = undefined, if x < 0 From the graph, the range is R = {y : -∞ < y < ∞} = (-∞, ∞)

The Product Property

One property of exponents is how the exponents add: bp bq = bp+q If we take the logarithm of both sides of that equation we get: logb(bp bq) = logb(bp+q) The right-hand side of this equation reduces to: logb(bp bq) ⩵ p + q Let’s let m = bp and n = bq, so we get: logb(m ·n) = p + q But, if we take the log of m and the log of n we get logb(m) = logb(bp) = p and logb(n) = logb(bq) = q This means that logb(m ·n) = logb(m) + logb(n)

The Quotient Property

Another property of exponents is: bp bq = bp-q If we repeat the process that we used for the Addition Property we can write: logb bp bq

= logb(bp-q) = p - q

Using the same definitions of m and n we get logb m n

= p - q

So logb m n

= logb(m) - logb(n)

The Power Property

Another property of exponents is:

(bp)q = bp·q

Again, let’s take the logs of both sides of this equation: logb((bp)q) = logb(bp·q) = p ·q Let’s also use the same definitions of m: m = bp As before: logb(m) = p So logb(mq) = p ·q = q · p And logb(mq) = q logb(m)

Change of Base

If you take a look at calculator, it (most likely) only has two log buttons: one for common logs (base 10) and

• ne for natural logs (base e). In some problems involving logs, the base is not 10 or e but you still need to

calculate its value. For example, log4(17) = ? To calculate this, we want to change the base from 4 to either 10 or e. In other words, we want to write our log expression into the form: log4(17) = k ·log(17)

• r j ·ln(17)

where j and k are “correction factors” to give us the right answer. Let’s do an example to see the pattern...

Change of Base Example

Evaluate: y = log4(17) Let’s get rid of the base 4 by using exponents: 4y = 4log4(17) = 17 Now lets take the common log of both sides: log(4y) = log(17) If we use the Power Property we get y log(4) = log(17) If we divide both sides by log(4) we get our answer: y = log4(17) = log(17) log(4)

= 2.0437

In this example, k = 1/log(4). Be sure to notice where the number 4 came from: it’s the original base.

Generalizing the Change of Base Rule

We can convert between any two bases. Let’s start with a logarithm with base b that we want to covert to a logarithm with base a: y = logb x As before, let’s first get rid of the base b: by = x Next, let’s take the logarithm of both sides (using base a): loga(by) = loga x Now apply the Power Property: y loga b = loga x Finally, we solve for y: y = loga x loga b So the Change of Base Rule for logarithms is: logb x = loga x loga b

Expanding Logarithmic Expressions

“Expanding” is one form of manipulation. Expansions are done using the properties of logarithms:

The Power Property

logb(mq) = q logb(m)

The Product Property

logb(m ·n) = logb(m) + logb(n)

The Quotient Property

logb m n

= logb(m) - logb(n)

Expanding with the Power Property

If we have a logarithmic expression such as log53 we can use the Power Property to expand the expression to log53 = 3 log(5) Here are more examples: log(5p) = p log(5) logg20 = 20 log(g) lnx-2 = -2 ln(x) log4(pq) = q log4(p)

Expanding with the Product Property

If we have a logarithmic expression such as log(100 x) we can use the Product Property to expand the expression to log(100 x) = log(100) + log(x) = log102 + log(x) = 2 + log(x)

Expanding with the Quotient Property

If we have a logarithmic expression such as log3 81 w we can use the Quotient Property to expand the expression to log3 81 w

= log3(81) - log3(w) = 4 - log3(w)

Expanding Radical Expressions

If the expression has a radical sign such as log2 x

4



the first step is to convert the radical to an exponent: log2 x

4

 = log2x1/4

and then use the Power Property: log2 x

4

 = log2x1/4 = 1

4 log2(x)

Expanding with Combinations of Properties

Some expansions require the use of more than one property. For example, logb x2 y3 z4 There is often more than one way to expand. In this example, it easiest to begin with the Quotient Property: logb x2 y3 z4

= logbx2 y3 - logbz4

We can now use the Product Property and the Power Property: logb x2 y3 z4

= logbx2 y3 - logbz4 = logbx2 + logby3 - 4 logb(z)

Finally, we use the Power Property again: logb x2 y3 z4

= 2 logb(x) + 3 logb(y) - 4 logb(z)

(Challenge) Another Combination Example

Sometimes it’s easier to simplify an expression before using the properties. For example, log x7 x5x2 - 2 x + 1 In one step, by using exponent properties, distribution and the Quotient Property, this can be written as logx7 - logx7 - 2 x6 + x51/2 We can finish the expansion using the Quotient Rule: 7 log(x) - 1 2 logx7 - 2 x6 + x5 Note that we cannot expand this any further (we cannot distribute the log function over addition or subtraction).

Condensing Logarithmic Expressions

The reverse of expanding a logarithmic expression is called condensing. In general, the order that we would have used to expand an expression is reversed in order to condense it.

The Power Property

q logb(m) = logb(mq)

The Product Property

logb(m) + logb(n) = logb(m ·n)

The Quotient Property

logb(m) - logb(n) = logb m n

Condensing with the Power Property

Let’s start with a simple example using the Power Property: 1 3 log(x) = logx1/3 While we could have converted the fractional exponent, 1/3, into a radical sign, that’s rarely needed.

Condensing with the Product Property

If we have log terms being added, we can condense using the Product Property log2(x) + log2(y) = log2(x y)

Condensing with the Quotient Property

If we have log terms being subtracted, we can condense using the Quotient Property log7(a) - log7(b) = log7 a b

Condensing with Combinations of Properties

Condensing often requires more than one operation and multiple properties. For example, 3 ln(x) + 2 ln(z) lnx3 + lnz2 lnx3 z2 Here is another example: 1 3 log3(11) - 1 2 log3(5) + log3(x) log3111/3 - log351/2 + log3(x) log3 111/3 x 51/2

(Challenge) Complicated Example

Here is a more complicated example: 1 5

(log3 x + log3 y - log3 z) - 2 log3(x - 5) - 4 log3 z - log3(16)

1 5 log3 x y z

• (2 log3(x - 5) + 4 log3 z + log3(16))

log3 x y z

1/5

• log3(x - 5)2 + log3 z4 + log3(16)

log3 x y z

1/5

• log316 (x - 5)2 z4

log3 x1/5 y1/5 16 z1/5 z4(x - 5)2 log3 x1/5 y1/5 16 z21/5(x - 5)2

(Challenge) Change of Base

Be careful when the logs have different bases, such as log3(x) + 2 log2(x) The first step is to change the base (typically to either common or natural log): log3(x) + 2 log2(x) = log(x) log(3)

+ 2 log(x)

log(2) We can now proceed to condense: logx1log(3) + logx2log(2) logx1log(3) x2log(2) log x

1 log(3) + 2 log(2)

(Challenge) Change of Base, continued

Note that we could have factored first: log3(x) + 2 log2(x) log(x) log(3)

+ 2 log(x)

log(2) 1 log(3)

+

2 log(2) log(x) log x

1 log(3) + 2 log(2)

(which is a much easier approach)

Solving Exponential Equations: Example 1

Let’s start with a simple equation: 42 x-3 = 8 Step 1: Take the logarithm of both sides: log42 x-3 = log(8) Step 2: Use properties of logs to expand

(2 x - 3) log(4) = log(8)

Step 3: Use algebra to isolate x

(2 x - 3) log(4) = log(8)

2 x - 3 = log(8) log(4) 2 x = log(8) log(4)

+ 3

x = log(8) 2 log(4)

+ 3

2

Solving Exponential Equations: Example 1, continued

At this point, we have an exact expression for x. As a challenge, we can attempt to simplify the expression: x = log(8) 2 log(4)

+ 3

2

=

log23 2 log22

+ 3

2

= 3 log(2)

4 log(2)

+ 3

2

= 3

4

+ 3

2

= 3

4

+ 6

4

= 9

4 It’s always a good idea to see if the answer makes sense by plugging back into the original equation: 42 x-3 = 42 (9/4)-3 = 49/2-6/2 = 43/2 = 23 = 8

Solving Exponential Equations: Example 2

Sometimes, we won’t get nice integers or fractions for answers... 34 x+2 = 5-2 x+1 Step 1: Take the log of both sides: ln34 x+2 = ln5-2 x+1 Step 2: Use properties of logs to expand:

(4 x + 2) ln(3) = (-2 x + 1) ln(5)

Step 3 : Use algebra to isolate x: 4 x ln(3) + 2 ln(3) = -2 x ln(5) + ln(5) 4 x ln(3) + 2 x ln(5) = ln(5) - 2 ln(3) 2 x(2 ln(3) + ln(5)) = ln(5) - 2 ln(3) x = ln(5) - 2 ln(3) 2 (2 ln(3) + ln(5))

Solving Exponential Equations: Example 2, continued

Once again we might be tempted to stop and use a calculator at this point, but it’s always worth trying to simplify as much as possible, even if it only makes using a calculator easier (and less error-prone) x = ln(5) - 2 ln(3) 2 (2 ln(3) + ln(5))

=

ln(5) - ln(9) 2 (ln(9) + ln(5))

=

ln(5 /9) ln(81) + ln(25)

=

ln(5 /9) ln(81 ·25) Using a calculator we get: x = -0.077204988 Again, it’s always a good idea to check the answer: 34 x+2 = 5-2 x+1 34 (-0.0772)+2 = 5-2 (-0.0772)+1 31.6912 = 51.1544 6.4107 = 6.4105

Solving Exponential Equations: Example 3

When the exponential equations uses e as a base, it is always best to solve using natural logarithms. 7 e2 x + 4 = 21 Step 1: Simplify as much as possible: 7 e2 x + 4 = 21 7 e2 x = 17 e2 x = 17 /7 Step 2: Take the log of both sides: lne2 x = ln(17 /7) Step 3: Solve for x: 2 x = ln(17 /7) x = ln(17 /7) 2

= 0.4436516

Solving Exponential Equations: Example 4 (Challenge)

When the exponential function has exponents with terms with powers greater than 1, it is possible for there to be more than one answer. Here’s an example: 5 ex2-2 = 30 Step 1: Simplify as much as possible: ex2-2 = 6 Step 2: Take the log of both sides: lnex2-2 = ln(6) x2 - 2 = ln(6) Step 3: Solve for x: x2 = 2 + ln(6) x = ± 2 + ln(6) = ±1.94724407

Solving Logarithmic Equations: Example 1

Let’s start with a simple equation: log(x) = 11 To solve, we use exponentiation: 10log(x) = 1011 x = 1011

Solving Logarithmic Equations: Example 2

In some logarithmic equations, we need to solve for the base: logb(243) = 5 One way to solve is to use exponentiation: blogb(243) = b5 b5 = 243 We can now use exponent properties to solve for b:

b51/5 = 2431/5

b = 2431/5 = 3

Solving Logarithmic Equations: Example 2, continued

Another way of solving this is to use the Change of Base Rule: logb(243) = 5 log(243) log(b)

= 5

log(b) = log(243) 5 Next, we use exponentiation to get b by itself: b = 10log(243)5 = 3 Sometimes, we can notice a nice simplification instead: log(b) = log(243) 5

=

log35 5

= 5 log(3)

5

= log(3)

Solving Logarithmic Equations: Example 3

When there are multiple terms, it is often useful to condense the logarithms first. Here is an example: ln(8 w) + ln(2 w) = 12 Step 1: Condense: ln16 w2 = 12 Step 2: Use exponentiation: eln16 w2 = e12 16 w2 = e12 Step 3: Solve for w: w2 = e12 16 w = ± e12 16

= ± e6

4

= ±100.8572

Solving Logarithmic Equations: Example 4

Sometimes we will encounter expressions as the argument of the log function. Here is an example: log4(8 x - 4) = 2 We solve these by “undoing” the log with an exponentiation and then simplifying: 4log4(8 x-4) = 42 8 x - 4 = 16 8 x = 20 x = 5 2

Solving Logarithmic Equations: Example 4

Here is a more complicated example: log2 p - log2(3 p - 2) = log2 9 Step 1: Condense: log2 p 3 p - 2

= log2 9

Step 2: Use exponentiation: p 3 p - 2

= 9

Step 3: Solve for p: p = 9 (3 p - 2) p = 27 p - 18 26 p = 18 p = 9 13