1. Logarithms and Logarithmic Functions
Chapter 4
Section 4.2
Logarithmic Functions and Models
Logarithmic Functions
We investigate logarithms, logarithmic functions and their connection to exponential functions.
This module was originally Section 5-4 of Chapter 5 from Gary Rockswold’s College Algebra with Modeling and Visulaizations. The current sequence of modules is from Robert Blitzer’s College Algebra, the current module being Section 4.2 of Chapter 4.
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2. Logarithmic Functions
Logarithms and Logarithmic Functions
What Is a Logarithm ?
Recall the exponential function ax
We define a new function, the logarithm with base a , as
where logax is the power of the base a
For example
with base a , a > 0 , a ≠ 1
loga x = y ay
= x
that equals x
What Is a Logarithm ?
We define a logarithm as a power of a base number and connect it directly to a corresponding exponential function with the same base number – that is, the logarithm is the exponent for the corresponding exponential base.
log232 = 5
since
25 = 32
and
log = 3
since
103 = 1000
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3. Logarithmic Functions
Logarithms and Logarithmic Functions
loga x = y
So we can think of the logarithm as an exponent
Since y IS the logarithm, we could also write
So, acting on variable x or y or … with either function ay
= x
alog ax
ay =
= x
logaay = logax
and
= y
What Is a Logarithm ?
We define a logarithm as a power of a base number and connect it directly to a corresponding exponential function with the same base number – that is, the logarithm is the exponent for the corresponding exponential base.
… and then on that result with the other function…
yields the original variable value
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4. Logarithmic Functions
Logarithms and Logarithmic Functions
Consider the exponential function f(x) = ax , 0 < a ≠ 1
We can show that f(x) is a 1–1 function
Hence f(x) does have an inverse function f–1(x)
f–1(f(x)) = f–1(ax) = x
Since we showed that
we call this inverse the logarithm function with base a
Since logax is an inverse of ax then
loga(ax) = x and
= a logax
logaay = logax
= y
Logarithmic Functions
Here we make the connection that loga x and ax are inverses of each other, so each of them is a 1-1 function.
We note that the domain of one is also the range of the other. The domain of ax is the set of real numbers R and its range is the set of all positive real numbers, (0, ∞).
The domain of the logarithm function is the set of all positive real numbers, (0, ∞) and its range is the set of all real numbers, R.
logax a
= x
Question:
What are the domain and range of ax ?
What are the domain and range of logax ?
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5. Logarithmic Functions
Logarithms and Logarithmic Functions
A Second Look
Graphs of f(x) = ax and f–1(x) = logax
are mirror images with respect
to the line y = x
Note that y = f–1(x) if and
only if f(y) = ay = x
y = logax iffi ay = x x y
f(x) = ax
, for a > 1
(1, a)
f–1(x) = logax ●
(0, 1) ● ●
(a, 1) ●
(1, 0)
y = x
Logarithmic Functions: A Second Look
We look at the graphs of f(x) = ax and f–1(x) = logax and verify the domains and ranges of the functions. We note that, as with all inverse function pairs, the graphs are symmetric with respect to the line y = x.
As the illustration shows, each point (x, y) on one graph has a corresponding point (y, x) on the other graph. This is exactly the concept of inverse function. The points (x, y) and (y, x) can also be viewed simply as ordered pairs in the sets of ordered pairs that are the functions.
Remember:
logax is the power of a that yields x
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6. Logarithmic Functions
Logarithms and Logarithmic Functions
A Third Look
loga x = y means ay = x
What if x = a ?
and ay = a
So for any base a > 0, a ≠ 1,
logaa = 1
What if x =1 ?
Then loga1 = y so ay = 1
and thus y = 0
So for any base a > 0 , a ≠ 1 ,
loga1 = 0
f–1(x) = logax x y ●
(0, 1)
y = x
f(x) = ax
(1, a)
, for a > 1
Then logaa = y
= a1
so that y = 1
WHY?
(a, 1)
(1, 0)
Logarithmic Functions: A Third Look
This illustrates a couple of basic facts about logarithms:
The log of the base is always 1
The log of 1 is always 0
These facts are true for any base.
Also note that ay = a1 implies y = 1 because the exponential function is The same reasoning applies to noting that ay = 1 implies that y = 0, because
a0 = 1 = ay and we again invoke the 1-1 property of exponential functions.
From inspection of the graphs of loga x and ax it is clear that the two functions are never equal for any value of x.
WHY?
Question:
Can logax = ax ?
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7. Logarithmic Functions
Logarithms and Logarithmic Functions
A Fourth Look
What if 0 < a < 1 ?
Then what is f–1(x) ?
Are the graphs still symmetric with respect to line y = x ?
YES !
Is it still true that
f(1) = a and f–1(x) = 1 ?
Is it possible that
loga x = ax ?
on the line y = x x y
f(x) = ax
, for a < 1
(a, 1) ● ●
(0, 1) ● ●
(1, a) ●
(1, 0)
y = x
Logarithmic Functions: A Fourth Look
This slide explores what happens when the base a is less than 1. The exponential growth function ax becomes an exponential decay function. The reflection of its graph through the line y = x is the graph of its inverse function, logax.
Clearly the two graphs intersect on the line y = x. The basic facts about both exponential and logarithmic functions still hold true:
The log of the base is always 1
The log of 1 is always 0
The exponential of 1 is always the base
The exponential of 0 is always 1
Since these facts are true for any base, they are certainly true for base a, where 0 < a < 1.
Having established that the graphs intersect on the line y = x the question we might ask is: for what value of x is this true? That is, can we solve the equation logax = ax
for x? We shall explore this question shortly.
f–1(x) = logax
Question:
For what x does logax = ax ?
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8. Logarithmic Functions
Logarithms and Logarithmic Functions
Example
Let a = ½
Then f(x) = ax
Then what is f–1(x) ?
f–1(x) = logax = log½x
It is still true that
f(1) = (½)1 = ½
and f–1(½) = log½(½) = 1
Where do the graphs intersect?
y = (½)x = log½x = x x y
f(x) = ax
, for a = ½
= (½)x
(½,1) ● ●
(0,1) ● ●
(1,½) ●
(1,0)
y = x
Logarithmic Functions: A Fourth Look Example
This slide explores what happens when the base is less than 1. The exponential growth function f(x) = ax becomes the exponential decay function f(x) = (½)x for a = ½ < 1. The inverse of this function is the logarithmic function f–1(x) = log½x, shown in the illustration.
Note that in this case both functions are decreasing functions that intersect the line y = x. This means that on that line the two function values are identical to the value of x chosen in each case. Since the inverse function must carry the value of y produced by one function back to the original value of x chosen for the other, the values of y must be identical when y = x.
To determine where f(x) = f –1(x) we must employ graphical and/or numerical methods, since there is no solution for logax = ax in closed form (i.e., no symbolic solution). By plotting the graphs of f(x), f–1(x), and y = x we can use a graphing calculator to trace a point along one of the graphs, say y = x, until we see that y = x. This should occur when y = x ≈
If you have a TI graphing calculator, setting the window size to [–.5, 2, 1] by [–.5, 2, 1] makes this fairly obvious.
f–1(x) = log1/2x
WHY ?
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9. Exponential/Logarithmic Comparisons
Logarithms and Logarithmic Functions
Compare ax with logax
Exponential Inverse Function
102 = 100
34 = 81
(½)–5 = 32
25 = 32 log232 = 5
5–3 = 1/125 log5(1/125) = –3
e3 ≈
Common Logarithm
Base 10
Log10 x = Log x
Natural Logarithm
Base e
Loge x = ln x
log10100 = 2
log381 = 4
log1/232 = –5
loge( )
≈ 3
Exponential - Logarithmic Comparisons
The Common Logarithm (base 10) is also known as the Briggsian logarithm after Henry Briggs (1561 – 1639) of Gresham College, London who computed the first table of common logarithms.
The Natural, or Naperian, Logarithm was invented by John Napier (1550 – 1617) Scottish laird (Baron of Murchiston) in a slightly different form. It was Napier who named this value the logarithm from the Greek words logos (ratio) and arithmos (number).
Napier did not think in terms of a base but in terms of a geometric sequence based on the numbers 1 – 10 – 7 . In practice the Naperian log L of a number N was given by the expression
Thus Naper’s logarithm of 107 is 0 and his logarithm of 107(1 – 1/107) is 1. Dividing his numbers by 107 yields very nearly a system of logarithms with base 1/e. The ratios in his system were geometric ratios of proportional distances measured along a pair lines. The modern version of the natural logarithm uses base e where The variable n can be replaced by 1/x as x approaches 0 , that is,
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e as n 
e as x
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10. One-to-One Property Review
Logarithms and Logarithmic Functions
Since ax and loga x are 1-1
1. If x = y then ax = ay
2. If ax = ay then x = y
3. If x = y then loga x = loga y
4. If log xa = loga y then x = y
These facts can be used to solve equations
Example: Solve 10x – 1 = 100
10x – 1 = 102
x – 1 = 2
x = 3
One-to-One Property Review
The salient feature to observe about 1-1 functions is that f(x) = f(y) if and only if x = y. The illustration shows what this means in terms of the exponential and logarithm functions.
We can use the fact that both the exponential and the logarithm functions are 1-1 to solve exponential and logarithm equations.
In the example we can move the variable from the exponent to a multiplier by using the logarithm to act on each side of the equation. This helps us form a linear equation in which we isolate x. By computing log 4 and log 10, in the example, we can solve for x. Note that log 10 = 1, since log without an explicit base is by convention log10. Solving for x gives us x =
Solution set :
{ 3 }
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11. Logarithms and Logarithmic Functions
Solving Equations
Logarithms and Logarithmic Functions
Solve
1. log3 x = –3
With inverse function: 2.
With definition:
3log x 3
= 3–3
3log x 3
= 3–3 27 1 = x
3–3 = x 27 1 = x
Solution set:
{ } 27 1 6  4 = x
log36
This is already solved for x , so simplify
Solving Equations
Solving a simple logarithm equation is done very much like solving an exponential equation: apply the inverse function to reduce the equation to a linear or polynomial equation.
In the first example we simply exponentiate both sides of the equation, which gets rid of the logarithm, leaving a simple linear equation to solve. The solution is then trivial.
The alternative solution simply applies the definition of logarithm directly, giving a simple linear equation, with the solution already isolated.
In the second example we again exponentiate both sides of the equation to produce a radical equation. Solving by raising both sides to the same power produces a simple exponential equation that is solved by applying the 1-1 property of exponential functions. 6  4
36x =
By definition the logarithm (i.e. x) is the
power of the base that yields 6  4
61/4
62x = 4 1 = 2x 8 1 = x
Solution set:
{ } 8 1
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12. Logarithms and Logarithmic Functions
Solving Equations
Logarithms and Logarithmic Functions
Solve
3. logx 5 = 4
By inverse function:
5 = x4
By definition: x
logx 5 = x4
( loga x = b iffi ab = x )
x4 = 5  4 x4 5 =
x =  5 4 =
x , for x > 0
–x , for x < 0 =
 x   5 4
{ }
Solution set:
Solving Equations
This example starts by exponentiating both sides, giving a trivial polynomial equation, which is solved by finding the fourth root of both sides.
The alternate solution applies the 1-1 property of logarithm functions, to produce the same power function.
Note that the solution can’t be negative, since that would force the logarithm base to be negative – which would violate the definition of logarithm ! Hence only the positive root can be considered.
Since x2 = 5 yields x =   5
Question:
, then why ?  5 4
doesn’t x4 = 5 yield x = 
If x =  5 –
Note:
then logx 5 is defined with a negative base !!
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13. Logarithms and Logarithmic Functions
Solving Equations
Logarithms and Logarithmic Functions
Solve
4. log3 (x2 + 5) = 2
By inverse function:
9 – 5 = x2
= 4x
log4 12 = log4 (4x)
By definition: = 3
log3 (x2 + 5) 32 =
x2 + 5
( 2 is the power of 3 that yields x2 + 5 )
32 = 9 = x2 + 5  4  = x
x2 = 4 = x 2 
x =  2
{  2 }
Solution set:
Solving Equations
In the first example we apply the inverse function and exponentiate both sides, producing a polynomial equation. The standard procedure is then followed to solve for x.
The alternative solution simply applies the definition of logarithm directly. Clearly the solutions are the same.
In the second example we again apply the inverse function, this time by finding the logarithm of both sides. However, we arrive at the embarrassing position of having to find the power of 4 that yields 12, i.e. log412. There is no closed-form solution for this equation. The standard approach to solving this equation is to identify the two sides of the equation as two separate functions: y1 = log412 and y2 = x and then approximate the intersection of the two graphs. OR, we try a change-of-base.
With modern calculators, such as the TI-83, TI-84, TI-85 and later models, choose the Math button and then the Solver option. Complete the equation 0 = with the terms of the given equation regrouped on the right. Press the ENTER key to see the result: x = approximately. The calculator has of course solved the equation as described above with iterated approximations. In the next section we will see how to change the base of the logarithm to something else, say log10 or common log. Then, the calculation is very simple:
= x
Question:
Now, how do we find log4 12 ?
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14. Logarithms and Logarithmic Functions
Solving Equations
Logarithms and Logarithmic Functions
Remember: ax and logax are inverses
1. To remove a variable from an exponent,
find the logarithm of the exponential form
2. To remove a variable from a logarithm,
exponentiate
Solving Equations
These are just review reminders about handling exponential and logarithmic equations.
The simple approach is to apply the inverse function in each of these cases, that is, apply the logarithm in the first case, and exponentiate in the second case.
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15. Logarithms and Logarithmic Functions
The Richter Scale
Logarithms and Logarithmic Functions
For fixed intensity x0 (as measured with a seismometer) the ratio of the seismic intensity x of an earthquake, relative to x0 , is measured by
This is the famous Richter Scale for measuring the relative “strength” of earthquakes
For x > x0 note that
R = log10 ( ) x x0 x x0
> 1 so
R = log10 ( ) x x0
> 0
The Richter Scale
This problem introduces the well known Richter scale for measuring earthquakes. The intensity of any earthquake is measured against a standard very “small” earthquake intensity, here represented by x0.
The “scale” measurement is actually the common logarithm of the ratio of a quake’s intensity to the standard intensity. Because the standard x0 is set so low, the intensity of any “real” earthquake x is much larger, meaning that the ratio x/x0 is always greater than 1. This gives positive Richter Scale values.
By the same means, we can compare any two earthquakes to see how much stronger (or weaker) one is relative to the other.
If two earthquakes have Richter scale readings of R1 and R2 with intensity values of x1 and x2, respectively, such that
then so that
that is, I2 = 10 I1. Thus a one-point increase in R corresponds to a ten-fold increase in intensity x. Conversely, a one-point decrease in R corresponds to a ten-fold decrease in intensity.
Question:
How fast does R grow ?
If R increases by 1 what is the change in x ?
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16. Richter Scale Comparisons
Logarithms and Logarithmic Functions
In 1992 the Landers earthquake produced a Richter scale value of 7.3 compared with the 1994 Northridge earthquake which hit 6.7 on the Richter scale
How much more powerful was the Landers earthquake, expressed as a ratio ?
Let RL = Landers intensity and RN = Northridge intensity
So RL = log10
From the definition of logarithm
Thus L = x and N = x0106.7
Now all we need is the ratio of L to N ... L x0
RN = log10 N x0
and
= 7.3
= 6.7
Richter Scale Comparisons
Here we compare two well publicized domestic earthquakes that occurred in 1992 and 1994 in California. The reported intensity readings were 7.3 and 6.7, respectively, from Northridge and from Landers.
A generally asked question was: how do we compare the relative strengths of these two earthquakes?
To answer this question, we need to understand that all Richer scale readings are relative to a standard small intensity x0. This represents a baseline seismometer reading against which all other actual seismometer readings are compared. This in turn allows us to compare actual earthquake readings without knowing the standard intensity value. Clearly, if any actual reading is smaller than x0 then the Richter scale value would be negative – which it has never been. So, x0 must be chosen to ensure that no measured intensity is ever less than x0. One way to ensure this is to choose x0 = 1.
If we were to choose x0 = 1, then
Rs = 4.8 = log10 (S/x0) = log10 S
so that
S = =
and all other intensities would then be larger than this value. L x0
107.3 =
and N x0
106.7 =
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17. The Richter Scale (continued)
Logarithms and Logarithmic Functions L N = x x
The ratio is
Thus L = 3.981N , i.e times as strong as Northridge
How much stronger was Landers than the smallest recorded earthquake with Rs = 4.8 ?
Landers was 316 times stronger than the smallest quake =
107.3 – 6.7 =
100.6
≈ 3.981 L S = x x =
107.3 – 4.8 =
102.5
≈ 316
The Richter Scale (continued)
Here we see how the intensity of different earthquakes compare. These are real earthquakes in recent California history. The calculations should be self explanatory.
One thing to note is that when one hears of a quake at 6.0 on the Richter scale this is not just twice the strength of a 3.0 quake – it is in fact it is 1000 times more intense, since quakes A and B have Richter Scale values given by and so that and
This is very clearly not a linear relationship. So, logarithmic scales are used to show a wide range of values using very small numbers.
Recent earthquakes in Haiti were reported by the media as 6.1 and 7.0 on the Richter scale, along with the comment that the smaller one was “nearly the same magnitude” as the larger. Is this true? Using the same calculations as in the illustration, the ratio of the first to the second comes down to
(Quake 1)/(Quake 2) = (107.0) / (106.1) ≈ 7.94
almost eight times larger. Is that “nearly the same” ?
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18. Earthquake Comparison Ratios
Logarithms and Logarithmic Functions
Richter Scale
2010 Haiti
1994 Northridge
1992 Landers
2010 Chile
2004 Indonesia
1960 Chile
Relative Strength Ratios
6.1
3.98
6.7
7.94
2.00
7.0
3.98
501.2
2.00
7.3
63.10 ?
31.62
199.53
Earthquake Comparisons
Richter Scale readings are often reported in the news media without much understanding of what they mean. Most people think of this as the often used “goodness” scale of 1 to 10, that is, a linear scale, which of course the Richter scale is not.
The readings shown in the slide are taken from news reports of earthquakes around the world in recent years. The “intensity” measured is related to the total amount of kinetic energy released as tectonic plates in the earth’s crust shift their relative positions, usually in a matter of a few seconds. This intensity is normally measured by seismic instruments called seismometers, which measure (typically) vertical movement of the earth at their locations.
The point to make here is that the Richter scale, being a logarithmic scale, shows differences in readings as ratios. That is, an increase anywhere on the scale of 0.3 yields an intensity twice as large as the lower value. Thus, the increase from 6.7 to 7.0 (Northridge to Haiti) shows a doubling of intensity, just as an increase from 7.0 to 7.3 (Haiti to Landers) is a doubling again. So, an increase from 6.7 to 7.6 would give an earthquake 8 times stronger than the 6.7 earthquake.
Just for perspective, the 9.3 Indonesian earthquake was approximately 200 times stronger than the 7.0 Haitian quake, and approximately 1585 times stronger than the 6.1 Haitian quake. The strongest quake ever recorded was a 9.5 quake in Chile in 1960, which was times the strength of the 6.1 Haitian quake.
8.8
100.0
3.16
9.3
1.58
9.5
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19. Earthquake Comparison Graph
Logarithms and Logarithmic Functions
4,000
3,000
2,000 R
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
1.0 x
1,000 ● ●
Chile 9.5 ●
Indonesia 9.3
Chile 8.8 20 x 10 R
8.0
6.0
7.0 ● ● ●
Landers 7.3
Haiti 7.0 ●
Landers 7.3 ●
Haiti 7.0
Earthquake Comparison Graph
The recent major quakes already listed are shown here on a scale showing their relative and absolute intensities. Note that very small changes in R can correlate with very large changes in intensity, measured with numbers in the billions.
A change in the Richter scale value of 1.0 corresponds to a factor of 10 times larger (or smaller) in magnitude. Thus a quake of magnitude 9.0 is 10 times stronger than one of magnitude 8.0, which in turn is 10 times stronger than a quake of magnitude Thus a difference of 2.0 on the Richter scale corresponds to a factor of 100 times in magnitude. ●
Northridge 6.7 ●
Haiti 6.1
Intensity x 106
Intensity x 106
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20. Earthquake Comparison Ratios
Logarithms and Logarithmic Functions
Richter Scale
2011 Hawaii
2010 Haiti
2010 Chile
2011 Japan
2004 Indonesia
1960 Chile
Relative Strength Ratios
4.5
79.43
6.4
316.2
398.1
3.981
7.0
31,622.8
63.10
8.8
100.0 ?
1.585
1,258.9
Earthquake Comparisons
The readings shown in the slide are taken from news reports of earthquakes around the world in recent years. The “intensity” measured is related to the total amount of kinetic energy released as tectonic plates in the earth’s crust shift their relative positions, usually in a matter of a few seconds. This intensity is normally measured by seismic instruments called seismometers, which measure (typically) vertical movement of the earth at their locations.
The point to make here is that the Richter scale, being a logarithmic scale, shows differences in readings as ratios. That is, an increase anywhere on the scale of 0.3 yields an intensity approximately twice as large as the lower value. It follows that an increase of another 0.3 on the scale yields another doubling of intensity, and hence an increase of 0.6 on the scale yields a fourfold increase in intensity. Thus, the increase from 6.4 to 7.0 (Hawaii in 2011 to Haiti in 2010) shows intensity increases by a factor of 3.981, or approximately fourfold.
Just for perspective, the 9.0 Japanese earthquake was approximately times stronger than the 6.4 Hawaiian quake which occurred on the same day, and approximately 31,622.8 times stronger than the 4.5 Hawaiian aftershock. The strongest quake ever recorded was a 9.5 quake in Chile in 1960, which was times the strength of the 6.4 Hawaiian quake and 100,000 time the strength of the 4.5 Hawaiian aftershock.
9.0
1.995
9.3
3.162
1.585
9.5
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21. Earthquake Comparison Ratios
Logarithms and Logarithmic Functions
Richter Scale
2011 Hawaii
1989 California
2010 Haiti
2010 Chile
2011 Japan
1964 Alaska
1960 Chile
Relative Strength Ratios
6.4
3.162
6.9
3.981
125.9
1.259
7.0
398.1
63.10
8.8
100.0 ?
1.585
398.1
Earthquake Comparisons
The readings shown in the slide are taken from news reports of earthquakes around the world in recent years. The “intensity” measured is related to the total amount of kinetic energy released as tectonic plates in the earth’s crust shift their relative positions, usually in a matter of a few seconds. This intensity is normally measured by seismic instruments called seismometers, which measure (typically) vertical movement of the earth at their locations.
The point to make here is that the Richter scale, being a logarithmic scale, shows differences in readings as ratios. That is, an increase anywhere on the scale of 0.3 yields an intensity approximately twice as large as the lower value. It follows that an increase of another 0.3 on the scale yields another doubling of intensity, and hence an increase of 0.6 on the scale yields a fourfold increase in intensity. Thus, the increase from 6.4 to 7.0 (Hawaii in 2011 to Haiti in 2010) shows intensity increases by a factor of 3.981, or approximately fourfold.
Just for perspective, the 9.0 Japanese earthquake was approximately times stronger than the 6.4 Hawaiian quake, but only times stronger than the 6.9 California quake. The strongest quake ever recorded was a 9.5 quake in Chile in 1960, which was times the strength of the 6.9 California quake and times the strength of the 6.4 Hawaiian quake.
9.0
1.585
9.2
3.162
1.995
9.5
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22. Logarithms and Logarithmic Functions
Think about it !
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