In this section we will be working with Properties of Logarithms in an ...

[Pages:10]16-week Lesson 33 (8-week Lesson 27)

Properties of Logarithms and Solving Log Equations (Part 2)

In this section we will be working with Properties of Logarithms in an attempt to take equations with more than one logarithm and condense them down into just a single logarithm.

Properties of Logarithms: a. Product Rule:

log() + log() = log( )

When two or more logarithms with the same base are added, those logarithms can be condensed into one logarithm whose argument is the product of the original arguments ( in the example above)

Order is not important when multiplying or adding, so changing the order of the factors in an argument or changing the order of the terms being added together does not change the answer.

b. Quotient Rule:

log() - log() = log ()

When two or more logarithms with the same base are subtracted, those logarithms can be condensed into one logarithm whose argument is the quotient of the original arguments ( )

Order is important when dividing and subtracting; the numerator is always the argument of the first log term listed (or the argument of the term listed first is always the numerator).

c. Power Rule: log() = log()

A factor times a logarithm can be re-written as the argument of the logarithm raised to the power of that factor

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16-week Lesson 33 (8-week Lesson 27)

Properties of Logarithms and Solving Log Equations (Part 2)

Example 1: Solve each of the following logarithmic equations and CHECK YOUR SOLUTIONS. LEAVE ANSWERS IN EXACT FORM, DO NOT APPROXIMATE.

a. 2 = log3() - log3( - 1)

b. A

2

=

log3

( )

-1

32 =

-1

9 =

-1

9( - 1) =

b. log2( + 7) + log2() = 3 log2(( + 7)()) = 3 log2(2 + 7) = 3 2 + 7 = 23 2 + 7 = 8

9 - 9 =

2 + 7 - 8 = 0

8 = 9

( + 8)( - 1) = 0

= 9

8

= -8 ; = 1

Regardless of what type of answer you come up with (negative, positive, or zero), you must check your answer to verify that it results in a positive argument. To do so, ALWAYS plug your answer back into the original equation.

Replacing

with

9 8

will

make

each argument in the original

equation

positive,

so

=

9 8

is a valid answer.

Replacing with -8 will make

each argument in the original equation negative, so = -8 is not a valid answer. = 1 is a valid answer because it makes the arguments positive.

=

=

2

16-week Lesson 33 (8-week Lesson 27)

Properties of Logarithms and Solving Log Equations (Part 2)

In Example 1, the Properties of Logarithms were only used to combine logarithms in each problem. This is how we will be using the Properties of Logarithms in this class, to combine logarithms in order to reduce the number logarithms we have to just one, so that we can then convert that one logarithm to exponential form to solve.

If an equation has more than one logarithm on either side of the equation, use the Properties of Logarithms to simplify as much as possible, then solve by converting to exponential form. Remember that you must check your answers when solving log equations to verify that they make the original arguments positive.

Example 2: Solve each of the following logarithmic equations and

CHECK YOUR SOLUTIONS. LEAVE ANSWERS IN EXACT

FORM, DO NOT APPROXIMATE.

a. 2 ln() - ln(5 - 6) = 0

2

=

log3

( )

-1

b. log2(1 - ) = 1 - log2( - 1) log2(1 - ) + log2( - 1) = 1

2

=

log3

( )

-1

log2((1 - )( - 1)) = 1

2

=

log3

( )

-1

2

=

log3

( )

-1

2

=

log3

( )

-1

2

=

log3

( )

-1

2

=

log3

( )

-1

2

=

log3

( )

-1

2 = log3 (-1)

2

=

log3

( )

-1

2

=

log3

( )

-1

log2( - 1 - 2 + ) = 1

The quadratic

log2(-2 + 2 - 1) = 1

equation I ended up with was

-2 + 2 - 1 = 21

not factorable,

so I chose

-2 + 2 - 1 = 2

to solve it by completing

-2 + 2 - 3 = 0

the square. When I did, I

2 - 2 + 3 = 0

ended up with an equation

2

-

2

+

(-22)2

=

-3

+

( ) -2

2 that is not

possible

2

using real numbers.

2 - 2 + 1 = -3 + 1

Therefore

the log

( - 1)2 = -2

equation given on

part b. has no real solutions.

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16-week Lesson 33 (8-week Lesson 27)

Properties of Logarithms and Solving Log Equations (Part 2)

b. a

c. 1 log( + 1) = log(1 - )

2

1

log( + 1)2 = log(1 - )

log( + 1) = log(1 - )

log( + 1) - log(1 - ) = 0

log (1-+1) = 0

+1 1-

=

100

+1 = 1

1-

+ 1 = 1 -

( + 1)2 = (1 - )2

+ 1 = (1 - )(1 - ) + 1 = 1 - 2 + 2 0 = 2 - 3

0 = ( - 3)

0 = ; - 3 = 0

= 0 ; = 3

This example demonstrates using two different Properties of Logarithms (the Power Rule first and the Quotient Rule second). Notice that when log( + 1) = log(1 - ), we could have simply set the arguments equal to each other and solved. In other words, if log( + 1) = log(1 - ), then + 1 = 1 - . You can see in the solution that we end up with this equation eventually, but only after using the Quotient Rule for Logarithms and converting from log form to exponential form. This shortcut can be used because logarithmic functions are one-to-one functions, so if log() = log(), then = .

You are welcome to use this shortcut or not, it makes no difference; your answers should be the same regardless. If you plan to use this shortcut, please keep in mind this only works when you have one logarithm equal to another. You cannot take this shortcut if you have any other terms or factors in the equation (such as Example 2 parts a. and b. on the previous page).

Replacing with 0 in the original equation makes each argument positive, so = 0 is a valid answer. Replacing with 3 results in a negative argument (1 - 3 = -2), so = 3 is not a valid answer.

=

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16-week Lesson 33 (8-week Lesson 27)

Properties of Logarithms and Solving Log Equations (Part 2)

Example 3: Solve each of the following logarithmic equations and CHECK YOUR SOLUTIONS. LEAVE ANSWERS IN EXACT FORM, DO NOT APPROXIMATE.

a. log( + 2) - log() = 2 log(4)

b. log4(3 + 2) = log4(5) + log4(3)

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16-week Lesson 33 (8-week Lesson 27)

c. ln( - 2) = 1 - ln()

Properties of Logarithms and Solving Log Equations (Part 2)

ln( - 2) + ln() = 1

ln(( - 2)()) = 1

log(2 - 2) = 1 2 - 2 = 1

2 - 2 + (-2)2 = + (-2)2

2

2

2 - 2 + 1 = + 1

( - 1)2 = + 1

- 1 = ? + 1

= 1 ? + 1

Notice that unlike on some of the previous problems, we cannot take the shortcut and simply set the arguments equal to each other, because there is no way to set the logarithms equal to each other. That's why I put the logarithms on the same side of the equation, and then used the Product Rule to combine them.

Also, once I converted from log form to exponential form, I ended up with a quadratic equation. But because that quadratic equation contains the natural number , I will not be able to solve by factoring, and using the quadratic formula will be difficult. So that's why I solved by completing the square.

Finally, always be sure to check your answers to see if they make the arguments in the original log equation positive. Remember that we always need to use the original equation we were given to check.

= 1 + + 1 ; = 1 - + 1

= 1 + + 1 is approximately 2.928285 ..., which means if we plug this back into the original equation for , we'll get positive arguments. This makes = 1 + + 1 a valid answer.

= 1 - + 1 is approximately -0.928285 ..., which means if we plug this back into the original equation for , we'll get negative arguments. This makes = 1 - + 1 an invalid answer.

Keep in mind that while it may be necessary to approximate an answer in order to check it, you should always input exact answers only, unless LON-CAPA states otherwise.

= + +

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16-week Lesson 33 (8-week Lesson 27)

Properties of Logarithms and Solving Log Equations (Part 2)

d. log5(10) - log5(50) = log5(0.5) + log5(4 + 6)

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16-week Lesson 33 (8-week Lesson 27)

Properties of Logarithms and Solving Log Equations (Part 2)

e.

2

log2(1

-

)

-

log2

(-

1)

+

5

=

7

-

log2(9)

log2(1

-

)2

-

log2

(-

1)

+

log2(9)

=

7

-

5

log2(1

-

)

-

log2

(-

1)

+

log2(9)

=

2

log2 (1--1) + log2(9) = 2

log2

((1

-

)

?

(-

1))

+

log2(9)

=

2

log2

((1

-

)

(-

))

1

+

log2(9)

=

2

log2(- + 2) + log2(9) = 2

Notice once again that because we have constant terms (5 and 7) that do not cancel out, we cannot use the shortcut of setting two logarithms with the same base equal to each other, and then simply setting the arguments equal to each other. That's why I put the logarithms on the same side of the equation, and then used the Properties of Logarithms to combine them.

log2((2 - )(9)) = 2 log2(92 - 9) = 2 92 - 9 = 22 92 - 9 = 4 92 - 9 - 4 = 0

Also, once I combined all the logarithms and then converted from log form to exponential form, I ended up with a quadratic equation. This time I was able to solve by factoring.

92 + 3 - 12 - 4 = 0

3(3 + 1) - 4(3 + 1) = 0

(3 + 1)(3 - 4) = 0

3 + 1 = 0 ; 3 - 4 = 0

= - 1 ; = 4

3

3

Finally, always be sure to check your answers to see if they make the arguments in the original log equation positive. Remember that we always need to use the original equation we were given to check.

Replacing with - 1 in the original equation makes each argument positive, so = - 1 is a valid answer.

3

3

Replacing

with

4 3

results

not

only

in

a

negative

argument

(-

1

4

),

but

also

a

negative

value

under

a

square

root

3

(1

-

43),

so

=

4 3

is

not

a

valid

answer.

= -

8

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