Derivation – Rules for Logarithms

[Pages:10]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: Example: Example: Example:

log10100 = 2; 102 = 100 log101000 = 3; 103 = 1000 log10 .001 = ?3; 10?3 = .001 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

= blogb 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

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10logab = ab ab = (10loga)(10logb)

(10loga)(10logb) = 10 loga + logb 10 logab = 10 loga + logb

- Given - Substitution - Mult Rule Exp. - 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

a/b

=

10log a 10log b

= 10log a

10log b

10log a- log b

10 = loga/b 10log a-log b

log a/b = log a ? log b

- Given - Substitution

- Div Rule Exp. - Transitive Prop. - 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.

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Another helpful rule in logarithms can be seen by raising them to a power.

We know that have

a = 10log a. If each side is raised to the power of n, we

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

loga bx = loga n xloga b = loga n

x = loga n

loga b

- Def of log - log of both sides - Power rule ? logs

- Div Prop. Equality

logbn =

loga n loga b

- Substitution

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

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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 Using the definition Multiplying Solving for x

log2(x + 8)(x ? 4) = 3 (x + 8)(x ? 4) = 23 x2 + 4x ? 32 = 8 x2 + 4x ? 40 = 0

Use Quadratic Formula a = 1, b = 4, c = ?40

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x = -(4) ? 42 - 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, Using the product and exp rules

log(x?2) + log(2x?3) = 2logx 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

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Example Solve for x, Using the quotient rule Using the definition

log 4x ? log4(x ? 1) = ?

x

log4 x -1 = ?

x =4?

x -1

x =2

x -1

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.

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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

Dividing by 100 Taking the ln of both sides

The growth rate approximates .25 per hour.

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450 = 100er6 4.5 = e6r ln(4.5) = lne6r ln(4.5) = 6r ln 4.5 = r

6 r = .2507

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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 Taking the ln of both sides

200 = 100e r6.5 2 = e r6.5 ln(2) = ln e r6.5

ln(2) = 6.5 r

ln(2) r = 6.5

The good news, now we know the rate.

Substituting those values into the original equation, A = Pert

36 ln(2)

A = 100 e 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.

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