Goals
We will
Remember properties of exponents
Discover properties of logarithms
Use these properties to solve equations
Properties of Exponents
Simplify the following expressions by writing with multiplication or
division.
x 2x 3 =
x3
=
x2
2x 2y
3x
3y
=
=
Scalar
Compare the results of the following log calculations to discover
the first property. Remember logb x = y is the same as b y = x .
1
Calculate log2 2, log2 22 , log2 23 , log2 2n
2
Calculate log2 2, 2 log2 2, 3 log2 2, n log2 2
Addition
Compare the results of the following log calculations to discover
the first property. Remember logb x = y is the same as b y = x . A
calculator may be useful for numeric approximations.
1
Calculate log2 (2 · 2), log2 (2 · 3), log2 (2 · 4), log2 (2 · 5)
2
Calculate (log2 2) + (log2 2), (log2 2) + (log2 3),
(log2 2) + (log2 4), (log2 2) + (log2 5),
Subtraction
Compare the results of the following log calculations to discover
the first property. Remember logb x = y is the same as b y = x . A
calculator may be useful for numeric approximations.
1
Calculate log2 (2/2), log2 (2/3), log2 (2/4), log2 (2/5)
2
Calculate (log2 2) − (log2 2), (log2 2) − (log2 3),
(log2 2) − (log2 4), (log2 2) − (log2 5),
Inverse
Remember logb x and b x are defined as inverses just like
subtraction is for addition. Answer the following questions based
on this idea.
1
2log2 x =
2
log2 (2x ) =
3
log3 (35x +1 ) =
Basic Technique
The following demonstrates how to use the inverse relationship of
logarithms and exponentials to solve an equation.
3(25x ) − 1 = 8.
3(25x ) − 1+1 = 8+1. inverse of subtraction
3(25x ) = 9.
1
1 5x
3(2 ) =
9. inverse of multiplication
3
3
25x = 3.
log2 (25x ) = log2 3. inverse of exponential
5x
1
5x
5
x
= log2 3.
1
=
log2 3. inverse of multiplication
5
log2 3
=
.
5
Basic Technique
The following demonstrates how to use the inverse relationship of
logarithms and exponentials to solve an equation.
3 log5 (x − 7) + 4 = 10.
3 log5 (x − 7) + 4−4 = 10−4.
3 log5 (x − 7) = 6.
6
3 log5 (x − 7)
=
.
3
3
log5 (x − 7) = 2.
x − 7 = 52 . inverse of log
x − 7+7 = 25+7.
x
= 32.
Simplify
The following demonstrates a simplification of logarithms that is
useful for some future problems.
q
log7
2
x2 + 2
x2 − 4
q
=
=
log7 2 + log7
log7 2 + log7
=
log7 2 +
=
log7 2 +
=
log7 2 +
=
log7 2 +
=
log7 2 +
1
2
1
2
1
2
1
2
1
2
x2 + 2
x2 − 4
x2 + 2
1/2
x2 − 4
log7
x2 + 2
x2
−4
2
2
log7 (x + 2) − log7 (x − 4)
2
log7 (x + 2) − log7 ((x − 2)(x + 2))
2
log7 (x + 2) − (log7 (x − 2) + log7 (x + 2))
2
log7 (x + 2) −
1
2
log7 (x − 2) −
1
2
log7 (x + 2)).