Download Derivation – Rules for Logarithms

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)
= b log
b
x
=x
=x
From that we can see the following if the base of the logs is 10 – common
logarithms:
10loga = a
10logb = b
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10logab = ab
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10logab = ab
ab = (10loga)(10logb)
(10loga)(10logb) = 10 loga + logb
So we can see the
- Given
- Substitution
- Mult Rule Exp.
10 logab = 10 loga + logb
- Transitive Prop.
log ab = log a + log b
- Exp Equation
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
10 log a
a/b = log b
10
10 log a
= 10 log a − log b
log b
10
So we can see
10loga/b = a/b
- Given
- Substitution
- Div Rule Exp.
10loga/b = 10 log a − log b
- Transitive Prop.
log a/b = log a – log b
- Exp Equation
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
log an = n log a
So we can see
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
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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
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
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x=
−(4) ± 4 2 − 4(1)(–40
2(1)
−4 ± 16 + 160
2
−4 ± 176
x=
2
−4 ± 13.2
x=
2
x = 4.6
x = −8.6
x=
x can not be –8.6 because we can not take the log of a negative number,
x = 4.6 is the solution.
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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,
log 4x – log4(x – 1) = ½
Using the quotient rule
log4
x
=½
x −1
Using the definition
x
x −1
=4½
€
€
€
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
450 = 100er6
Dividing by 100
4.5 = e6r
Taking the ln of both sides
ln(4.5) = lne6r
ln(4.5) = 6r
ln 4.5
=r
6
r = .2507
The growth rate approximates .25 per hour.
€
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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
r=
ln(2)
6.5
The good news, now we know the rate.
Substituting those values into the original equation,€
A = 100 e
Using your calculator
A = Pert
36 ln(2)
6.5
A ≈ 4647.7 or 4648 bacteria
€ problem, you were not given
The formula is pretty straight forward. In this
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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The A = Pert formula is related to the compound interest formula when the
interest is being compounded continuously. A = P(1+
r nt
)
n
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.
r
A = P(1+ ) nt
n
Substituting those values in
.04 12⋅ 5
)
12
.04 60
10,000 = P(1+
)
12
10,000
=P
.04 60
(1+
)
12
10,000 = P(1+
€
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
) when
n
compounded annually, semi-annually, quarterly, monthly, weekly, daily,
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hourly, by the minute, by the second, etc. If we did all that, and we won’t,
we’d find (1+
1 n
) gets closer and closer to 2.71828247… . Rather than
n
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
)
x0
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.
M(x) = log(
Substituting
M(x) = log(
€
x
)
x0
7943
)
.001
M(x) = 6.9
€
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