3.4 Logarithms & Log Scales

3.4 Logarithms & Log Scales

This first page is just meant to give you an idea of how logarithms show up in various applications. It is informational only and will not be tested.

1. Why use logarithm functions?

• Consider linear scales in the first three graphs below versus a log scale in the fourth graph.

Mercury

Venus Earth

Mars

Jupiter

0 500 1000

Planet Distance from the SunHmillions kmL

Mars Jupiter

Saturn

Uranus Neptune Pluto

0 2 4 6

Planet Distance from the SunHbillions kmL

Pluto Alpha Centauri

0 20 40 60

Celestial Distance from the SunHtrillions kmL

Mercury

Jupiter

Pluto Alpha Centauri Andromeda Galaxy

Quasars

100 ₁₀2 ₁₀4 ₁₀6 ₁₀8 ₁₀10 ₁₀12 ₁₀14 ₁₀16 Celestial Distance from the SunHmillions kmL

In the first picture, we try to get a sense of the distance between planets in the solar system, using a linear scale of millions of km. Unfortunately, this only lets us get out to Jupiter on this scale.

In the second picture, we try using a linear scale of billions of km. Unfortu-nately, this starts to bunch up the inner planets near 0 on this scale, but lets us get out to Pluto.

In the third picture, we try using a linear scale of trillions of km. Unfortu-nately, this really bunches up everything at 0 on this scale, and leaves only one distinct point, Alpha Centauri.

In the fourth picture, we use a logarithm scale on millions of km. This really spreads out everything on this scale, as the distances between axis

hash marks grows, not by a constant (linearly), but constant increase by exponent (logarithmically).

• pH Scale: If [H+] is the hydrogen-ion concentration in gram-ions per liter, then pH =−log[H+].

Simply an example of how log base 10 is used in chemistry.

• Decibel Voltage Gain: IfEO is the output voltage of a device andEI is the input voltage,

the decibel voltage gain (dB gain) is

dB gain = 10 log

EO EI

.

An example of how log base 10 is used in acoustics.

• Isothermal Expansion: If the temperatureT is constant, the energyErequired to increase the volume of 1 mole of gas from an initial volumeVi to a final volume Vf is

E =RTln

Vf Vi

,

whereE is in joules, T in kelvins, and R= 8.314 joules per mole per kelvin.

An illustration of how the natural log (log base e) is used in science.

2. Properties of Logs (a) log_b1 = 0

For natural log, ln 1 = 0. This just says x= 1 is the x-intercept for all loga-rithms.

(b) log_bb = 1

For natural log, lne= 1. This is the inverse property, since the base of the log equals the base of the number plugged into the log.

(c) log_bbx =x

For natural log, lnex _{= x. Inverse property when the bases are equal.}

(d) blogbx =x

For natural log, elnx _{= x. Inverse property, since the base of the exponential} equals the base of the log.

(e) Expand (Split Apart) a Product, or Combine an Addition: log_b(xy) = log_bx+ log_by, specifically ln(xy) = lnx+ lny for base e.

If you have a log of a product of two things, you can split that into the sum of two logs. Conversely, if you have the sum of two logs, you can combine them into a single log of their product. It is used both ways.

(f) Expand (Split Apart) a Quotient, or Combine a Subtraction: log_b x

y = logbx−logby,

specifically lnx

y = lnx−lny.

If you have a log of a quotient of two things, you can split that into the difference of two logs. Conversely, if you have the difference of two logs, you can combine them into a single log of their quotient. It is used both ways.

(g) Power Rule: log_bxk_=klog

bx, specifically lnxk =klnx

If you have a log of a thing raised to a power, you can pull the power down in front of the log. Conversely, if you have a number in front of a log, you can raise that number into the power. It is used both ways.

(h) One-to-one function: log_bx= log_by⇐⇒x=y, specifically lnx= lny ⇐⇒x=y

Since the log is a one-to-one function (it passes the horizontal line test), you can cancel the logs from both sides, in some sense. Be careful (warning!): You must simplify both sides until there is one log on the left and one log on the right. Use the power rule to get numbers in front of logs into the power, and combine logs using parts (e) and (f ) to reduce each side of the equation to single logs.

(i) Change of Base: Change a logarithm base b to a logarithm base a, or base 10, base e. log_bx = logax

log_ab

= logx logb

= lnx lnb

They are all equal conversions! Since calculators only handle base 10 (log) and base e (ln), this formula allows us to change from any one base b to any other base a, including base 10 and/or base e to allow for calculation.

(j) Switching from a log expression to an exponential expression (and vice versa):

y= log_bx ⇐⇒ by =blogbx

⇐⇒ by =x

y= lnx ⇐⇒ ey =elnx ⇐⇒ ey =x

Solving forx above: it is inside a logarithm, so we convert to an exponential, use inverse rule (d) above, and get x in terms of y.

x=by ⇐⇒ log_bx= log_bby ⇐⇒ log_bx=y

x=ey ⇐⇒ lnx= lney ⇐⇒ lnx=y

Solving for the exponent y above: it is in the power, so we convert to a logarithm, and get y in terms of x.

(k) NOT A RULE!

logx= logy+ logz 6=⇒ x=y+z

Again, we cannot just cancel logs willy-nilly. As mentioned before in (h), we must simplify both sides until there is one log on the left and one log on the right:

logx = logy+ logz

= log(yz) means

logx = log(yz),

so actually,

x= yz after applying log property (h).

3. Examples:

(a) Express in terms of logs of a, b, c: log₂ a 3_b2

√ c

log₂ a 3_b2 √

c = log2(a

3_b2_)−log 2

√ c

= log₂a3 + log₂b2 −log₂c1/2

= 3 log₂a+ 2 log₂b− 1

2log2c.

First, we split the log up using earlier properties (f ) and (e) above. Then we use the power rule (g) to bring the powers down front.

(b) Write as a single logarithm: 5 log₄(x−7)−3 log₄x

5 log₄(x−7)−3 log₄x = log₄(x−7)5 −log₄x3

= log₄ (x−7) 5

x3 .

Reversing the steps of the previous example, we first move the numbers in front of the logs into the powers. Then we combine the difference of two logs into a single quotient, using property (f ).

(c) Express in terms of logs of a, b, c: ln √3bc_a

ln bc

3

√

a = lnbc−ln

3

√ a

= lnb+ lnc−lna1/3

= lnb+ lnc− 1 3lna.

Split the log up using properties (f ) and (e) above, and then use the power rule (g) to pull the powers down front.

(d) Write as a single logarithm: 3 lna+1₂lnb−4 ln(c2+ 1)

3 lna+ 1

2lnb−4 ln(c

2 _{+ 1) = lna}3 _{+ lnb}1/2 _−ln(c2_{+ 1)}4

= lna3b1/2 −ln(c2 + 1)4

= ln a

3√_b (c2_{+ 1)}4.

Move the numbers in front of the logs into the powers. Then we combine the difference of two logs into a single quotient, using property (f ).

(e) Simplify log₄2 + log₄32

log₄2 + log₄32 = log₄(2∗32)

= log₄(64)

= log₄43

= 3.

First, combine the logs using property (e) above. Then, to facilitate using inverse property (c) above, realize that 64 can be written as an exponential using a base of 4.

(f) Simplify log₂80−log₂5

log₂80−log₂5 = log₂ 80 5

= log₂16

= log₂24

= 4.

First, combine the difference of two logs using property (f ) above, and sim-plify. Then, realize that 16 can be written as an exponential base 2.

(g) Simplify −₃1 log₂8

−1

3 log28 = −1

3 log22 3

= −1

3 ∗3

= −1.

First, realize that 8 can be written as an exponential base 2, since 23 = 8. Then, use inverse property (c) above to see log₂23 _{= 3. Lastly, simplify!.}

(h) Change log₈5 to a natural logarithm

log₈5 = ln 5_{ln 8} = 0.77398 using the change of base idea.

(i) Change log₉20 to a base 10 logarithm

log₉20 = log 20_{log 9} = 1.3634 using the change-of-base idea again.

(j) Simplify (log₂5)(log₅7) by a change of base

(log₂5)(log₅7) = (log₂5)

log₂7 log₂5

= (log₂5)

_log

27 log₂5

= log₂7 or,

(log₂5)(log₅7) =

_log

55 log₅2

(log₅7)

= (1)

_log

57 log₅2

= log₂7.

Since the two bases of the logs are different, we must change one to the other through the change-of-base formula.

(k) Solve forf: lnf = lnf0−cln(t+ 1)

First, we move the c up to the power using the power rule, and then combine the difference of two logs into one log of a quotient. Then, use the one-to-one property of logs to cancel the logs as in property (h) above:

lnf = lnf0 −cln(t+ 1)

= lnf0 −ln(t+ 1)c

= ln f0

so lnf = ln f0

(t+1)c, and thus

f = f0 (t+ 1)c.

(l) Solve for P: logP = logc−klogW

First, move the k up to the power using the power rule, and then combine the difference of two logs into one log of a quotient. Then, use the one-to-one property of logs to cancel the logs as in property (h) above:

logP = logc−klogW

= logc−logWk

= log c

Wk, so logP = log _Wck, and thus

P = c Wk.

4. Richter Scale for Earthquakes: Let M be the magnitude of the earthquake,I be the intensity of the shock waves, and I0 be the intensity of the smallest quake that can be measured on a seismograph (based on the distance of the machine from the epicenter). Then

M = log

I I0

.

The Richter Scale gives a magnitude number, once the intensity is measured by a machine.

(a) How much more intense is an 8.5 earthquake on the Richter scale than a 6.8?

M1 = 8.5 = log (I1/I0) =⇒ 108.5 =I1/I0. M2 = 6.8 = log (I2/I0) =⇒ 106.8 =I2/I0. Therefore,

I1 I2

= 10 8.5_I

0 106.8_I

0

= 101.7 ≈ 50.

An 8.5 earthquake is about 50 times more intense than a 6.8 earthquake!

(b) How much more intense was the 8.9 earthquake in Tohoku, Japan on March 11, 2011 than the 8.0 earthquake in Sichuan, China in May 2008?

M1 = 8.9 = log (I1/I0) =⇒ 108.9 =I1/I0.

Therefore,

I1 I2

= 10 8.9_I

0 108.0_I

0

= 100.9 ≈ 7.9.

An 8.9 earthquake is about 8 times more intense than an 8.0 earthquake. Summary for parts (a) and (b): Given Richter scale magnitudes M1 and M2, the change in intensity is

10M1−M2_.

(c) The earthquake that hit Haiti in January 2010 had a magnitude of 7.0. What would be the magnitude of an earthquake 20 times more intense?

Reverse the steps of parts (a) and (b) just above. Let the unknown Richter magnitude be x. Then,

20 = I1 I2 =

10xI0

107.0_I0 = 10 x−7.0

.

To solve for x, take log of both sides:

log 20 = x−7 so