Download Solve Logarithmic Equations

Solve Logarithmic Equations
Solving an equation means determining the value of the unknown variable for a
given function value. To solve a logarithmic equation, use the following strategy:
1. Isolate the logarithm if the logarithm is not alone on one side
2. Change the logarithm to exponential form
3a. Write both sides of the equation with the same base
4a. Set the exponents equal to each other and solve
OR
3b. Simplify the exponential term
4b. Use algebra to solve for the variable
Example 1: Solve: log2 64 = x
2x = 64
2x = 26
x=6
Change the logarithm to exponential form
Rewrite with the same base, 64 = 26
Set the exponents equal to each other
Example 2: Solve: log125 5 = x
125x = 5
(53)x = 51
53x = 51
3x = 1
x=
1
3
Change the logarithm to exponential form
Rewrite with the same base, 125 = 53 and 5 = 51
Use the Power Rule for exponents
Set the exponents equal to each other
Divide by 3
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010; and Intermediate
Algebra, by Andrew Gloag, Anne Gloag, and Mara Landers, CK-12 Foundation, CC-BY 2013. Licensed
under a Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/
by/3.0).
Example 3: Solve: log3
3x =
1
27
=x
1
27
3x = 3-3
x = -3
Change the logarithm to exponential form
Rewrite with the same base,
1
27
=
1
33
= 3-3
Set the exponents equal to each other
Example 4: Solve: log5 x = 2
52 = x
25 = x
Change the logarithm to exponential form
Simplify the exponential term
Example 5: Solve: log2 (3x + 5) = 4
24 = 3x + 5
16 = 3x + 5
11 = 3x
11
=x
3
Change the logarithm to exponential form
Simplify the exponential term
Subtract 5 from both sides
Divide by 3
Example 6: Solve: 1 + 2 log3 (x – 5) = 7
1 + 2 log3 (x – 5) = 7 Isolate the logarithm
2 log3 (x – 5) = 6 Subtract 6 from both sides
log3 (x – 5) = 3 Divide by 2
33 = x – 5
27 = x – 5
32 = x
Change the logarithm to exponential form
Simplify the exponential term
Add 5 to both sides
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010; and Intermediate
Algebra, by Andrew Gloag, Anne Gloag, and Mara Landers, CK-12 Foundation, CC-BY 2013. Licensed
under a Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/
by/3.0).
We can also solve equations in which both sides of the equation contain logs. If
the logarithms have the same base, the arguments of the logarithm must be equal.
Example 7: Solve: log2 (3x – 1) = log2 (5x – 7)
3x – 1 = 5x – 7
-2x – 1 = -7
-2x = -6
x=3
Bases are the same so the arguments are equal
Subtract 5x from both sides
Add 1 to both sides
Divide by -2
Example 8: Solve: log5 (9x) = log5 (3x + 8)
9x = 3x + 8
6x = 8
x=
x=
8
6
4
3
Bases are the same so the arguments are equal
Subtract 3x from both sides
Divide by 6
Reduce the fraction
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010; and Intermediate
Algebra, by Andrew Gloag, Anne Gloag, and Mara Landers, CK-12 Foundation, CC-BY 2013. Licensed
under a Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/
by/3.0).