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MATH 107
Section 4.5
Logarithmic and
Exponential Equations
2© 2010 Pearson Education, Inc. All rights reserved
EXAMPLE 1 Solving an Exponential Equation
a. 25x
= 125
Solve each equation.
b. 9x
= 3x+1
Solution
3© 2010 Pearson Education, Inc. All rights reserved
OBJECTIVE Solve
exponential equations when
both sides are not expressed
with the same base.
Step 1 Isolate the exponential
expression on one side of the
equation.
Step 2 Take the common or
natural logarithm of both
sides.
Step 3 Use the power rule,
loga M r
= r loga M.
EXAMPLE 2
Solving Exponential Equations Using the
Logarithms
EXAMPLE Solve for x:
5 ∙ 2x – 3
= 17.
4© 2010 Pearson Education, Inc. All rights reserved
OBJECTIVE Solve
exponential equations when
both sides are not expressed
with the same base.
Step 4 Solve for the variable.
EXAMPLE 2
Solving Exponential Equations Using the
Logarithms
EXAMPLE Solve for x:
5 ∙ 2x – 3
= 17.
Practice Problem
5© 2010 Pearson Education, Inc. All rights reserved
6© 2010 Pearson Education, Inc. All rights reserved
EXAMPLE 3
Solving an Exponential Equation with
Different Bases
( ) ( )
( )
2 3 1
ln5 ln3
ln5 ln3
2 ln5 3ln5 ln3 ln3
2 ln5 ln3 ln3 3ln5
2ln5 ln3 ln3 3ln5
ln3 3ln5
2.795
2ln5
3
n3
2 1
l
x x
x x
x x
x
x
x x
− +
=
=
− = +
− = +
− = +
+
= ≈
−
− +
Solve the equation 52x–3
= 3x+1
and approximate
the answer to three decimal places.
When
different
bases are
involved,
begin with
Step 2.
Solution
Practice Problem
7© 2010 Pearson Education, Inc. All rights reserved
8© 2010 Pearson Education, Inc. All rights reserved
SOLVING LOGARITHMS EQUATIONS
4
2log means 2 64 1x x= = =
Equations that contain terms of the form log a x
are called logarithmic equations.
To solve a logarithmic equation we write it in
the equivalent exponential form.
log2 x = 4 log3 2x −1( ) = log3 x + 2( )
log2 x − 3( ) + log2 x − 4( ) = 1
9© 2010 Pearson Education, Inc. All rights reserved
EXAMPLE 7 Solving a Logarithmic Equation
Solve: 4 + 3log2 x = 1.
We must check our solution.
Solution
2
2
2
1
4 3log 1
3log 1 4 3
log 1
2
1
2
x
x
x
x
x
−
+ =
= − = −
= −
=
=
10© 2010 Pearson Education, Inc. All rights reserved
EXAMPLE 7 Solving a Logarithmic Equation
Solution continued
1
2






.The solution set is
1
2
.Check x =
?
2
?
2
?
2
?
?
1
4 3log 1
1
4 3log 1
2
4 3log 1
4 3
2
1
1 1
x
−
+ =
+ =
+ =
− =
=
Practice Problem
11© 2010 Pearson Education, Inc. All rights reserved
12© 2010 Pearson Education, Inc. All rights reserved
EXAMPLE 9 Using the Product and Quotient Rules
Solve: ( ) ( )2 2a. log 3 log 4 1x x− + − =
( ) ( )2 2b. log 4 log 3 1.x x+ − + =
Solution
( ) ( )
( ) ( )
( ) ( )
( ) ( )
2 2
2
1
2
2
a. log 3 log 4 1
log 3 4 1
3 4 2
7 12 2
7 10 0
2 5 0
x x
x x
x x
x x
x x
x x
− + − =
 − − = 
− − =
− + =
− + =
− − =
13© 2010 Pearson Education, Inc. All rights reserved
EXAMPLE 9 Using the Product and Quotient Rules
Solution continued
x − 2 = 0 or x − 5 = 0
x = 2 or x = 5
Check x = 2
Logarithms are not defined for negative
numbers, so x = 2 is not a solution.
( ) ( )
( ) ( )
2 2
2 2
log 3 log 4 1
log 1 lo
2 2
g 2 1
− + − =
− + − =
?
?
14© 2010 Pearson Education, Inc. All rights reserved
EXAMPLE 9 Using the Product and Quotient Rules
Solution continued
The solution set is {5}.
Check x = 5
( ) ( )
( ) ( )
2 2
2 2
log 3 log 4 1
log 2 log 1 1
1 0 1
1 1
5 5− + − =
+ =
+ =
=
?
?
?
15© 2010 Pearson Education, Inc. All rights reserved
EXAMPLE 9 Using the Product and Quotient Rules
Solution continued
b.
16© 2010 Pearson Education, Inc. All rights reserved
EXAMPLE 9 Using the Product and Quotient Rules
Solution continued
Check x = −2
1 1=
Check x = −5
log2 (−1) and log2 (−2) are undefined, so
solution set is {−
2}.
Practice Problem
17© 2010 Pearson Education, Inc. All rights reserved

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Lecture 12 sections 4.5 logarithmic equations

  • 1. MATH 107 Section 4.5 Logarithmic and Exponential Equations
  • 2. 2© 2010 Pearson Education, Inc. All rights reserved EXAMPLE 1 Solving an Exponential Equation a. 25x = 125 Solve each equation. b. 9x = 3x+1 Solution
  • 3. 3© 2010 Pearson Education, Inc. All rights reserved OBJECTIVE Solve exponential equations when both sides are not expressed with the same base. Step 1 Isolate the exponential expression on one side of the equation. Step 2 Take the common or natural logarithm of both sides. Step 3 Use the power rule, loga M r = r loga M. EXAMPLE 2 Solving Exponential Equations Using the Logarithms EXAMPLE Solve for x: 5 ∙ 2x – 3 = 17.
  • 4. 4© 2010 Pearson Education, Inc. All rights reserved OBJECTIVE Solve exponential equations when both sides are not expressed with the same base. Step 4 Solve for the variable. EXAMPLE 2 Solving Exponential Equations Using the Logarithms EXAMPLE Solve for x: 5 ∙ 2x – 3 = 17.
  • 5. Practice Problem 5© 2010 Pearson Education, Inc. All rights reserved
  • 6. 6© 2010 Pearson Education, Inc. All rights reserved EXAMPLE 3 Solving an Exponential Equation with Different Bases ( ) ( ) ( ) 2 3 1 ln5 ln3 ln5 ln3 2 ln5 3ln5 ln3 ln3 2 ln5 ln3 ln3 3ln5 2ln5 ln3 ln3 3ln5 ln3 3ln5 2.795 2ln5 3 n3 2 1 l x x x x x x x x x x − + = = − = + − = + − = + + = ≈ − − + Solve the equation 52x–3 = 3x+1 and approximate the answer to three decimal places. When different bases are involved, begin with Step 2. Solution
  • 7. Practice Problem 7© 2010 Pearson Education, Inc. All rights reserved
  • 8. 8© 2010 Pearson Education, Inc. All rights reserved SOLVING LOGARITHMS EQUATIONS 4 2log means 2 64 1x x= = = Equations that contain terms of the form log a x are called logarithmic equations. To solve a logarithmic equation we write it in the equivalent exponential form. log2 x = 4 log3 2x −1( ) = log3 x + 2( ) log2 x − 3( ) + log2 x − 4( ) = 1
  • 9. 9© 2010 Pearson Education, Inc. All rights reserved EXAMPLE 7 Solving a Logarithmic Equation Solve: 4 + 3log2 x = 1. We must check our solution. Solution 2 2 2 1 4 3log 1 3log 1 4 3 log 1 2 1 2 x x x x x − + = = − = − = − = =
  • 10. 10© 2010 Pearson Education, Inc. All rights reserved EXAMPLE 7 Solving a Logarithmic Equation Solution continued 1 2       .The solution set is 1 2 .Check x = ? 2 ? 2 ? 2 ? ? 1 4 3log 1 1 4 3log 1 2 4 3log 1 4 3 2 1 1 1 x − + = + = + = − = =
  • 11. Practice Problem 11© 2010 Pearson Education, Inc. All rights reserved
  • 12. 12© 2010 Pearson Education, Inc. All rights reserved EXAMPLE 9 Using the Product and Quotient Rules Solve: ( ) ( )2 2a. log 3 log 4 1x x− + − = ( ) ( )2 2b. log 4 log 3 1.x x+ − + = Solution ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 2 a. log 3 log 4 1 log 3 4 1 3 4 2 7 12 2 7 10 0 2 5 0 x x x x x x x x x x x x − + − =  − − =  − − = − + = − + = − − =
  • 13. 13© 2010 Pearson Education, Inc. All rights reserved EXAMPLE 9 Using the Product and Quotient Rules Solution continued x − 2 = 0 or x − 5 = 0 x = 2 or x = 5 Check x = 2 Logarithms are not defined for negative numbers, so x = 2 is not a solution. ( ) ( ) ( ) ( ) 2 2 2 2 log 3 log 4 1 log 1 lo 2 2 g 2 1 − + − = − + − = ? ?
  • 14. 14© 2010 Pearson Education, Inc. All rights reserved EXAMPLE 9 Using the Product and Quotient Rules Solution continued The solution set is {5}. Check x = 5 ( ) ( ) ( ) ( ) 2 2 2 2 log 3 log 4 1 log 2 log 1 1 1 0 1 1 1 5 5− + − = + = + = = ? ? ?
  • 15. 15© 2010 Pearson Education, Inc. All rights reserved EXAMPLE 9 Using the Product and Quotient Rules Solution continued b.
  • 16. 16© 2010 Pearson Education, Inc. All rights reserved EXAMPLE 9 Using the Product and Quotient Rules Solution continued Check x = −2 1 1= Check x = −5 log2 (−1) and log2 (−2) are undefined, so solution set is {− 2}.
  • 17. Practice Problem 17© 2010 Pearson Education, Inc. All rights reserved