This lecture discusses evaluating logarithmic expressions, rules of
logarithms, and solving logarithmic and exponential equations.
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Remember that in the discussion of logarithmic functions, an important
relation was provided to arrive at the inverse of an exponential function,
which is a logarithmic function: x = b^y if and only y=logb(x).
This relation allows an equation in exponential form to be rewritten it in
logarithmic form, and vice versa.
Example:
In this example, the exponential equation, 5^3 = 125, needs to be rewritten
as a log equation.
Look at the relation in the statement above. The initial equation is of the form
b^y = x. So b, the base, is 5; the exponent, y, is equal to 3, and the result,x,
is 125.
Now, write the logarithmic form according to the definition: log5(125) =3.
Pay very close attention to where everything goes in the logarithmic form:
the base in the exponential equation is the base of the logarithm; the
exponent becomes what the logarithm is equal to.
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Example:
This example starts out with the logarithmic form. Remember that the base
of the logarithm is the base of the exponent in the exponential equation. The
number on the right side of the equation represents the power to which the
base is raised. Finally, 1/16 represents the result of 2^-4. So 2^-4 = 1/16,
which makes sense, according to the rules of exponents.
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Example:
In part a, if the value of the expression is not know, set this expression equal
to an unknown and then rewrite it as an equivalent exponential equation:
12^x = 144.
12^x = 12^2, which means that x = 2.
So log 12(144) = 2.
In part b,
8^x = 1
Remember from the rules of exponents that b^0 =1, b =/= 0
So x = 0, which means that
Log8(1) = 0.
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In part c, log3(1/27). So it is necessary to find the power to which 3 must be
raised in order to obtain 1/27. To do this, think back to the rules of
exponents. The rule for negative exponents states that b^-m = 1/b^m. So 3^3 = 1/27.
This means that log3(1/27) = -3, since 3^-3 = 1/27.
In part d, log64(8). At first glance, the temptation is to say that this
expression is equal to 2, since 8^2 =64. But the base of this logarithmic
function is 64, not 8. So it is necessary to find a value that represents the
exponent to which 64 must be raised in order to obtain 8. Using rational
exponents: nth root of b = b^1/n, the sqrt(64) = 64 ^1/2 = 8. So log64(8) = ½.
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There are rules for evaluating logarithmic expressions, and they follow from
the rules of exponential expressions.
The Product Rule for Logarithms states that a single logarithm of a product
can be rewritten as the sum of logarithms of the individual terms in the
product. Notice that this rule does not say the following: logb(M) + logb(N) =
logb(M+N).
Based upon the Product Rule, it is easy to see that the Quotient rule states
that the logarithm of a quotient can be rewritten as the difference of the
individual terms in the quotient. Again, be careful in understanding what the
rule does not say: Logb(M) – logb(N) = logb(M –N)
Third, the Power Rule. This is an interesting and very useful rule. The Power
rule states that log of an exponential expression can be rewritten as the
product of the exponent of the exponential expression and the logarithm of
the base of the exponential expression.
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Two important logarithms are the common logarithm and the natural
logarithm. A common logarithm is a logarithm with a base of 10. The base 10
logarithm is written simply as log(x). The natural logarithm has the constant
e as its base. The natural logarithm has a unique notation; it is written as
ln(x). These two bases are the only two bases that can be evaluated on
most calculators.
All scientific and graphing calculators will have keys or functions that are
labeled ln and log. To evaluate a logarithm of another base using a
calculator, it is necessary to use the Change-of-Base formula. There are a
variety of forms of the change of base formula. This form takes any base
logarithm and rewrites it in terms of the natural logarithm: Logb(M) =
ln(M)/ln(b)
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Example:
In part a, log(x+1) + log(y) – log(6z). Rather than trying to tackle the whole
expression at one time, begin with the first two terms. The sum of the first
two terms can be rewritten using the Product Rule for logarithms. This sum
becomes a single logarithm going in reverse order of the Product Rule:
Log(y(x+1)) – log(6z)
Next, simplify this product slightly: log(xy+y) – log (6z). The result is having
the difference of two log terms with the same base, which looks like the right
side of the Quotient Rule for logarithms. So this expression can be rewritten
as the logarithm of a quotient: log(xy+y/6z)
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Example:
In part b, begin by writing each term as the logarithm of an exponential
expression, where the power of each exponential expression is the constant
that was multiplied times each logarithm: log3(a^4) – log3((ab)^5)
Next use rules of exponents to simplify the exponential expression of the
second term: log3(a^4) – log 3(a^5b^5)
Now the expression can be rewritten using the Quotient Rule:
log3(a^4/a^5b^5)
Finish by simplifying the quotient: log3(1/ab^5)
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Example:
In part c, this expression, like the last example, has a term that must be
rewritten using the Power Rule. Rewriting the second term, results in: ln(2t)
– ln((4u)^3) + ln(t+3) which simplifies to ln(2t)-ln(64u^3) + ln(t+3).
Next consider the difference of the first two terms, which using the Quotient
Rule can be rewritten as: ln(2t/64u^3) + ln(t+3) = ln(t/32u^3) + ln(t+3)
Now use the Product Rule to arrive at a single logarithm: ln(t/32t^3 * t+3) =
ln(t^2+3t/32u^3)
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To solve a logarithmic equation, rewrite the equation so that it is in the form
logb(x) = y. When it is in this form, the equation can be changed to its
equivalent exponential form, b^y = x.
After solving a logarithmic equation, be sure to check all potential solutions
in the ORIGINAL EQUATION. There are two things to watch out for when
checking a solution in the original equation: the base must be between 0 and
1 or must be greater than 1.
Note, you cannot evaluate the logarithm of 0 or a negative value.
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Example:
In part a, log2(x+4) = 5. This equation is already in the form logb(x) = y, so
begin by rewriting this equation in exponential form: 2^5 = x+4.
Next simplify and then solve for x:
32 = x+4
28 = x.
Remember to check this solution in the original equation:
Log2(28+4)=5
This solution checks, and so x =28 is the solution.
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Example:
In part b, logx(4) + logx(9) = 2. Begin by using the Rules of Logarithms to
simplify the left-hand side of this equation, specifically the Product Rule:
logx(36) = 2.
Next rewrite the equation as an equivalent exponential equation: x^2 = 36.
The two values of x that satisfy this equation are x =6 and x= -6. Check each
of these solutions in the original equation:
x = 6: log6(4) + log6(9) = 2.
Each of the log terms in this equation can be evaluated, and this solution
checks.
x = -6: log-6(4) + log-6(9) = 2. The log terms cannot be evaluated, since the
base is negative for each. So x = -6 is not a solution of this equation.
Thus, the sole solution is x = 6.
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Example:
In part c, there are logarithmic expressions on both sides of the equation, so
begin by using the Quotient Rule on the sum of terms on the left side: ln(x^24x) = ln(21)
Now, move the term on the right side to the left side: ln(x^2-4x) – ln(21) = 0.
Use the quotient rule to combine the two terms on the left: ln((x^2-4x)/21) =
0
Rewrite this equation as an exponential equation:
e^0 = (x^2-4x)/21
e^0 = 1, so 1 = (x^2-4x)/21
Multiply both sides by 21: 21 = x^2 -4x.
Since the result is a quadratic equation, collect all terms on the right side
and factor:
0 = x^2 -4x -21
0 = (x -7)(x+3)
x -7 = 0 and x+3 =0, which gives two potential solutions, x =7 and x = -3.
Notice that if x =-3, the two logarithms cannot be evaluated. So x = -3 is not
a solution to this equation.
x = 7 is the sole solution.
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When solving exponential equations was originally discussed, each of the
equations could be solved by rewriting the exponential terms in the equation
as powers of a common base. This cannot always easily be done.
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The natural log function may be applied to both sides of the equation. Note
that any base logarithmic function can also be applied, including the
common logarithm function.
Once this function has been applied, the Power rule can be used to rewrite
the equation and then solve for the variable.
The reason for using the natural logarithm or the common logarithm is that
each provides a decimal approximation of the solution, which is especially
useful in application problems.
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Example:
Apply the natural logarithm to both sides of the equation: ln 2^x = ln10.
Now use the Power Rule to rewrite the left side of this equation:
xln2 = ln10.
Note that ln2 is a number. So the left side is simply the product of a variable
and a number. Thus divide by this number to isolate x: x = ln10/ln2.
This expression cannot be further simplified. It represents the exact value of
the solution.
Using a calculator, ln10/ln2 is approximately 3.322.
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Example:
Clearly, it is impossible to write 25 and 7 as powers of a common base. So
apply the natural logarithm function to both sides of this equation:
ln(25^(x+2)) = ln(7^x).
Here, the Power Rule can be applied to both sides of the equation:
(x+2)ln(25) = xln(7). Notice the parentheses around x+2, which was the
expression in the exponent of the term inside of the logarithm.
Next use the distributive property to rewrite the left side: xln25 + 2ln25 =
xln7.
Then collect all terms containing the variable on the same side: 2ln25 =
xln7 – x ln25.
The terms on the right side are not like terms, and thus cannot be combined.
But, the common factor of x can be factored out: 2ln25 = x(ln7 – ln25)
ln7 – ln25 is a real number and is not equal to zero. So divide by this value
to isolate the variable: x = 2ln(25)/(ln7-ln25).
Again this is the exact value of the solution which is approximately equal to 5.057.
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Example:
This is an example where an exponential function is used to model the
growth of a population. Since the growth constant is positive, the population
is increasing on the island.
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Example:
Remember, the growth function is P(t) = 8.1e^0.019t. The initial population
size in this model is 8.1 million. So doubling this value, results in P = 16.2
million. Thus, it is necessary to solve for t when P =16.2 million.
Begin by dividing by 8.1 so that the exponential expression containing t is
only on the left side: 2 = e^0.019t
Next apply the natural log function to both sides of this equation: ln2 =
(0.019t)lne. Note that lne = loge(e) = 1.
So ln2 = 0.019t.
Dividing both sides by 0.019 results in ln2/0.019 = t.
It would not make much sense to tell a person that it will take ln2/0.019
years, so a decimal approximation is necessary. It will take approximately 36
years.
So 36 years after 1975 would be the year 2006.
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