Derivation – Rules for Logarithms

214753

Hanlonmath 800.218.5482 bill@hanlonmath.com 1 Derivation - Rules for Logarithms For all a > 0, there is a unique real number n such that a = 10n. The exponent n is called the logarithm of a to the base 10, written log10a = n. In general, the logba = n if and only if a = bn Example: log10100 = 2;  102 = 100 Example: log101000 = 3; 103 = 1000 Example: log10 .001 = -3; 10-3 = .001 Example: log525 = 2;  52 = 25 Since by1 = by2 iff y1 = y2. That implies that logbx1 = logbx2 iff x1 = x2 The inverse of the exponential equation, y = bx is found by interchanging the domain and range, the x and y. So the inverse of y = bx is x = by which is written as y = logbx. Rewriting that in functional notation, we have f(x) = bx and f-1(x) = logbx. We also know that f[f-1(x)] = x. Let's use that information and make some substitutions: f(x) = bx f[f-1(x)] = b

f -1 (x) = x = b log b x

= x From that we can see the following if the base of the logs is 10 - common logarithms: 10loga = a 10logb = b 10logab = ab

Hanlonmath 800.218.5482 bill@hanlonmath.com 2 10logab = ab - Given ab = (10loga)(10logb) - Substitution (10loga)(10logb) = 10 loga + logb - Mult Rule Exp. 10 logab = 10 loga + logb - Transitive Prop. log ab = log a + log b - Exp Equation So we can see the log ab = log a + log b. Therefore we can say, to find the logarithm of a product of positive numbers, you add the logarithms of the numbers. We can use a similar derivation to find the log a/b. Again we know 10loga = a 10logb = b 10loga/b = a/b 10loga/b = a/b - Given a/b =10

loga 10 logb - Substitution 10 loga 10 logb = 10 loga-logb - Div Rule Exp. 10loga/b = 10 loga-logb

- Transitive Prop. log a/b = log a - log b - Exp Equation So we can see log a/b = log a - log b Therefore we can say, to find the logarithm of a quotient of positive numbers, you subtract the logarithms of the numbers.

Hanlonmath 800.218.5482 bill@hanlonmath.com 3 Another helpful rule in logarithms can be seen by raising them to a power. We know that a = 10log a. If each side is raised to the power of n, we have an = (10 loga)n (10 loga)n = 10nlog a So we can see log an = n log a Therefore we can say, to find the logarithm of a power, you multiply the logarithm by the exponent. Sometimes it is helpful to change the base of a logarithm such as logbn to a logarithm in base. Let x = logbn bx = n - Def of log loga bx = loga n - log of both sides xloga b = loga n - Power rule - logs x = log

a n log a b - Div Prop. Equality logbn = log a n log a b

- Substitution So we can see to change the base of a logarithm, we have logbn = logan/logab

Hanlonmath 800.218.5482 bill@hanlonmath.com 4 Now that we have those rules, we can use them to solve equations. Essentially, there are two types of logarithmic equations; a log equals a number or a log = log. If a log equals a number, we use the definition. If a log equals a log, we drop the logs. Our initial job is to rewrite the exponential or logarithmic equations into one of those two forms using the rules we derived. Example Solve for x, log x + log (x-3) = 1 Using the product rule log x(x-3) = 1 Using the definition and knowing when a base is not written it is understood to be 10, we have x(x-3) = 101 Using the D-Prop x2 - 3x = 10 Solving for x x2 - 3x - 10 = 0 (x+2)(x-5) = 0 x + 2 = 0 or x - 5 =0 x = - 2 or x = 5 ***Important, you must check your answers! You can only take a log of a positive number. If x = -2, then we would be taking a log of a negative number - that can not be a solution. The answer is x = 5 Example Solve for x, log2(x + 8) + log2(x - 4) = 3 Using the product rule log2(x + 8)(x - 4) = 3 Using the definition (x + 8)(x - 4) = 23 Multiplying x2 + 4x - 32 = 8 Solving for x x2 + 4x - 40 = 0 Use Quadratic Formula a = 1, b = 4, c = -40

Hanlonmath 800.218.5482 bill@hanlonmath.com 5 €

x= -(4)±4 2 -4(1)(-40 2(1) x= -4±16+160 2 x= -4±176 2 x= -4±13.2 2 x=4.6 x=-8.6

x can not be -8.6 because we can not take the log of a negative number, x = 4.6 is the solution. Now, even though this problem took more steps to solve, that should not equate to this problem being more difficult. Rather than solving it by factoring and using the Zero Product Property, we used the Quadratic Formula. Now, let's look at a problem where we have logs on both sides of the equation. Remember the rule, once the equation is in simplified form, we drop the logs. Example Solve for x, log(x-2) + log(2x-3) = 2logx Using the product and exp rules log (x-2)(2x-3) = logx2 (x - 2)(2x - 3) = x2 2x2 - 7x + 6 = x2 x2 - 7x + 6 = 0 (x - 6)(x - 1) = 0 x = 6 or x = 1 When x = 1, results in taking a log of a negative number - can't happen! x = 6

Hanlonmath 800.218.5482 bill@hanlonmath.com 6 Example Solve for x, log 4x - log4(x - 1) = ½ Using the quotient rule log4 €

x x-1 = ½ Using the definition € x x-1 = 4 ½ € x x-1

= 2 2(x - 1) = x 2x - 2 = x x = 2 Checking the answer, x = 2 works. So you need to remember, there are two types of logarithmic problems; log equals number and log = log. If log = #, then we use logba = n if and only if a = bn and solve If log = log, then we use logbx1 = logbx2 iff x1 = x2 and solve But to use those, you first have to simplify the logarithmic expressions using the product, quotient, power or change of base rules.

Hanlonmath 800.218.5482 bill@hanlonmath.com 7 Many exponential word problems involving growth or decay are often described using A = Pert and solved using logarithms. A - end result of whatever you are looking for P - is the initial amount you are working with r - is the rate of growth or decay t - is the time To solve problems of growth or decay, you need to know that formula and the rest is easy - just substitute. Example A student places 100 bacteria into a petri dish. Six hours later, he measures 450 bacteria. Assuming exponential growth, what is the growth rate "r" for the bacteria? P = 100, A = 450 and t = 6 Substituting those values in the formula A = Pert 450 = 100er6 Dividing by 100 4.5 = e6r Taking the ln of both sides ln(4.5) = lne6r ln(4.5) = 6r €

ln4.5 6 =r r = .2507 The growth rate approximates .25 per hour.

Hanlonmath 800.218.5482 bill@hanlonmath.com 8 Example A certain bacteria doubles in population every 6.5 hours. Given that there were approximately 100 bacteria to start with, how many bacteria will there be in a day and a half? Using the formula A = Pert We are looking for A, P = 100, t = 36 (day and a half), r = ? When we substitute these values into the equation, we see we have two unknowns. In order to solve this problem, we need to know the rate. We can determine that by using the equation A = Pert knowing that the bacteria doubles every 6.5 hours. Substituting A = 200, P = 100, and t = 6.5, we have 200 = 100e r6.5 2 = e r6.5 Taking the ln of both sides ln(2) = ln e r6.5 ln(2) = 6.5 r r = €

ln(2) 6.5

The good news, now we know the rate. Substituting those values into the original equation, A = Pert A = 100 €

e

36ln(2)

6.5

Using your calculator €

A≈

4647.7 or 4648 bacteria The formula is pretty straight forward. In this problem, you were not given the rate explicitly. We had to use extra information in the problem to find it. The arithmetic would is cumbersome, using a calculator is a must.

Hanlonmath 800.218.5482 bill@hanlonmath.com 9 The A = Pert formula is related to the compound interest formula when the interest is being compounded continuously. €

A=P(1+

r n nt

In this formula: A - ending amount P - beginning amount n - number of compounding per year t - time in years Example Suppose that you plan to need $10,000 in five years when your nephew begins high school. You want to invest in investment program yielding 4% interest, compounded monthly. How much should you invest? A = 10,000, r = 4% or .04, t = 5, n= 12 since it is being compounded monthly. Substituting those values in €

A=P(1+

r n nt

10,000=P(1+

.04 12

12⋅5

10,000=P(1+

.04 12 60

10,000

(1+ .04 12 60
=P

To do the rest, break out the calculator. Now, we indicated these two formulas are related. Let's see how. By letting the principal, rate and years equal to one, we could find €

(1+ 1 n n when compounded annually, semi-annually, quarterly, monthly, weekly, daily,

Hanlonmath 800.218.5482 bill@hanlonmath.com 10 hourly, by the minute, by the second, etc. If we did all that, and we won't, we'd find €

(1+ 1 n n

gets closer and closer to 2.71828247... . Rather than writing out that number, we call it, you guessed it, e. Logarithms are useful in solving such problems as the magnitude of an earthquake. The Richter Scale is a common method to measure earthquakes. The scale converts seismographic readings into numbers that make it easier to understand. €

M(x)=log(

x x 0

M - magnitude of the earthquake x - intensity of the earthquake measure seismographically in mm x0 - zero level earthquake whose seismographic reading measures .001 at a distance of 100 km from the epicenter Example Find the magnitude of an earthquake given the seismographic reading of 7943 mm was recorded 100 km from the epicenter. Substituting €

M(x)=log(

x x 0

M(x)=log(

7943
.001

M(x)=6.9

Hanlonmath 800.218.5482 bill@hanlonmath.com 1 Derivation - Rules for Logarithms For all a > 0, there is a unique real number n such that a = 10n. The exponent n is called the logarithm of a to the base 10, written log10a = n. In general, the logba = n if and only if a = bn Example: log10100 = 2;  102 = 100 Example: log101000 = 3; 103 = 1000 Example: log10 .001 = -3; 10-3 = .001 Example: log525 = 2;  52 = 25 Since by1 = by2 iff y1 = y2. That implies that logbx1 = logbx2 iff x1 = x2 The inverse of the exponential equation, y = bx is found by interchanging the domain and range, the x and y. So the inverse of y = bx is x = by which is written as y = logbx. Rewriting that in functional notation, we have f(x) = bx and f-1(x) = logbx. We also know that f[f-1(x)] = x. Let's use that information and make some substitutions: f(x) = bx f[f-1(x)] = b

f -1 (x) = x = b log b x

= x From that we can see the following if the base of the logs is 10 - common logarithms: 10loga = a 10logb = b 10logab = ab

Hanlonmath 800.218.5482 bill@hanlonmath.com 2 10logab = ab - Given ab = (10loga)(10logb) - Substitution (10loga)(10logb) = 10 loga + logb - Mult Rule Exp. 10 logab = 10 loga + logb - Transitive Prop. log ab = log a + log b - Exp Equation So we can see the log ab = log a + log b. Therefore we can say, to find the logarithm of a product of positive numbers, you add the logarithms of the numbers. We can use a similar derivation to find the log a/b. Again we know 10loga = a 10logb = b 10loga/b = a/b 10loga/b = a/b - Given a/b =10

loga 10 logb - Substitution 10 loga 10 logb = 10 loga-logb - Div Rule Exp. 10loga/b = 10 loga-logb

- Transitive Prop. log a/b = log a - log b - Exp Equation So we can see log a/b = log a - log b Therefore we can say, to find the logarithm of a quotient of positive numbers, you subtract the logarithms of the numbers.

Hanlonmath 800.218.5482 bill@hanlonmath.com 3 Another helpful rule in logarithms can be seen by raising them to a power. We know that a = 10log a. If each side is raised to the power of n, we have an = (10 loga)n (10 loga)n = 10nlog a So we can see log an = n log a Therefore we can say, to find the logarithm of a power, you multiply the logarithm by the exponent. Sometimes it is helpful to change the base of a logarithm such as logbn to a logarithm in base. Let x = logbn bx = n - Def of log loga bx = loga n - log of both sides xloga b = loga n - Power rule - logs x = log

a n log a b - Div Prop. Equality logbn = log a n log a b

- Substitution So we can see to change the base of a logarithm, we have logbn = logan/logab

Hanlonmath 800.218.5482 bill@hanlonmath.com 4 Now that we have those rules, we can use them to solve equations. Essentially, there are two types of logarithmic equations; a log equals a number or a log = log. If a log equals a number, we use the definition. If a log equals a log, we drop the logs. Our initial job is to rewrite the exponential or logarithmic equations into one of those two forms using the rules we derived. Example Solve for x, log x + log (x-3) = 1 Using the product rule log x(x-3) = 1 Using the definition and knowing when a base is not written it is understood to be 10, we have x(x-3) = 101 Using the D-Prop x2 - 3x = 10 Solving for x x2 - 3x - 10 = 0 (x+2)(x-5) = 0 x + 2 = 0 or x - 5 =0 x = - 2 or x = 5 ***Important, you must check your answers! You can only take a log of a positive number. If x = -2, then we would be taking a log of a negative number - that can not be a solution. The answer is x = 5 Example Solve for x, log2(x + 8) + log2(x - 4) = 3 Using the product rule log2(x + 8)(x - 4) = 3 Using the definition (x + 8)(x - 4) = 23 Multiplying x2 + 4x - 32 = 8 Solving for x x2 + 4x - 40 = 0 Use Quadratic Formula a = 1, b = 4, c = -40

Hanlonmath 800.218.5482 bill@hanlonmath.com 5 €

x= -(4)±4 2 -4(1)(-40 2(1) x= -4±16+160 2 x= -4±176 2 x= -4±13.2 2 x=4.6 x=-8.6

x can not be -8.6 because we can not take the log of a negative number, x = 4.6 is the solution. Now, even though this problem took more steps to solve, that should not equate to this problem being more difficult. Rather than solving it by factoring and using the Zero Product Property, we used the Quadratic Formula. Now, let's look at a problem where we have logs on both sides of the equation. Remember the rule, once the equation is in simplified form, we drop the logs. Example Solve for x, log(x-2) + log(2x-3) = 2logx Using the product and exp rules log (x-2)(2x-3) = logx2 (x - 2)(2x - 3) = x2 2x2 - 7x + 6 = x2 x2 - 7x + 6 = 0 (x - 6)(x - 1) = 0 x = 6 or x = 1 When x = 1, results in taking a log of a negative number - can't happen! x = 6

Hanlonmath 800.218.5482 bill@hanlonmath.com 6 Example Solve for x, log 4x - log4(x - 1) = ½ Using the quotient rule log4 €

x x-1 = ½ Using the definition € x x-1 = 4 ½ € x x-1

= 2 2(x - 1) = x 2x - 2 = x x = 2 Checking the answer, x = 2 works. So you need to remember, there are two types of logarithmic problems; log equals number and log = log. If log = #, then we use logba = n if and only if a = bn and solve If log = log, then we use logbx1 = logbx2 iff x1 = x2 and solve But to use those, you first have to simplify the logarithmic expressions using the product, quotient, power or change of base rules.

Hanlonmath 800.218.5482 bill@hanlonmath.com 7 Many exponential word problems involving growth or decay are often described using A = Pert and solved using logarithms. A - end result of whatever you are looking for P - is the initial amount you are working with r - is the rate of growth or decay t - is the time To solve problems of growth or decay, you need to know that formula and the rest is easy - just substitute. Example A student places 100 bacteria into a petri dish. Six hours later, he measures 450 bacteria. Assuming exponential growth, what is the growth rate "r" for the bacteria? P = 100, A = 450 and t = 6 Substituting those values in the formula A = Pert 450 = 100er6 Dividing by 100 4.5 = e6r Taking the ln of both sides ln(4.5) = lne6r ln(4.5) = 6r €

ln4.5 6 =r r = .2507 The growth rate approximates .25 per hour.

Hanlonmath 800.218.5482 bill@hanlonmath.com 8 Example A certain bacteria doubles in population every 6.5 hours. Given that there were approximately 100 bacteria to start with, how many bacteria will there be in a day and a half? Using the formula A = Pert We are looking for A, P = 100, t = 36 (day and a half), r = ? When we substitute these values into the equation, we see we have two unknowns. In order to solve this problem, we need to know the rate. We can determine that by using the equation A = Pert knowing that the bacteria doubles every 6.5 hours. Substituting A = 200, P = 100, and t = 6.5, we have 200 = 100e r6.5 2 = e r6.5 Taking the ln of both sides ln(2) = ln e r6.5 ln(2) = 6.5 r r = €

ln(2) 6.5

The good news, now we know the rate. Substituting those values into the original equation, A = Pert A = 100 €

e

36ln(2)

6.5

Using your calculator €

A≈

4647.7 or 4648 bacteria The formula is pretty straight forward. In this problem, you were not given the rate explicitly. We had to use extra information in the problem to find it. The arithmetic would is cumbersome, using a calculator is a must.

Hanlonmath 800.218.5482 bill@hanlonmath.com 9 The A = Pert formula is related to the compound interest formula when the interest is being compounded continuously. €

A=P(1+

r n nt

In this formula: A - ending amount P - beginning amount n - number of compounding per year t - time in years Example Suppose that you plan to need $10,000 in five years when your nephew begins high school. You want to invest in investment program yielding 4% interest, compounded monthly. How much should you invest? A = 10,000, r = 4% or .04, t = 5, n= 12 since it is being compounded monthly. Substituting those values in €

A=P(1+

r n nt

10,000=P(1+

.04 12

12⋅5

10,000=P(1+

.04 12 60

10,000

(1+ .04 12 60
=P

To do the rest, break out the calculator. Now, we indicated these two formulas are related. Let's see how. By letting the principal, rate and years equal to one, we could find €

(1+ 1 n n when compounded annually, semi-annually, quarterly, monthly, weekly, daily,

Hanlonmath 800.218.5482 bill@hanlonmath.com 10 hourly, by the minute, by the second, etc. If we did all that, and we won't, we'd find €

(1+ 1 n n

gets closer and closer to 2.71828247... . Rather than writing out that number, we call it, you guessed it, e. Logarithms are useful in solving such problems as the magnitude of an earthquake. The Richter Scale is a common method to measure earthquakes. The scale converts seismographic readings into numbers that make it easier to understand. €

M(x)=log(

x x 0

M - magnitude of the earthquake x - intensity of the earthquake measure seismographically in mm x0 - zero level earthquake whose seismographic reading measures .001 at a distance of 100 km from the epicenter Example Find the magnitude of an earthquake given the seismographic reading of 7943 mm was recorded 100 km from the epicenter. Substituting €

M(x)=log(

x x 0

M(x)=log(

7943
.001

M(x)=6.9