Logarithms Logarithmic and Exponential Form

1 This instructional aid was prepared by the Tallahassee Community College Learning Commons. Logarithms A logarithm of a given number , is the exponent required for the base , to be raised to in order to produce that number . log = = Note that means " " Logarithmic and Exponential Form Change logarithm equations to Exponential form or Exponential equations to Logarithmic form using the definition of a logarithm. Example: Given 432 =8 , change the equation to Logarithmic form. Solution: Compare the equation to the definition and rewrite it.

2 Definition: log = = Given: 432 =8 Notice that =4, =8,and =32,respectively. Therefore, using the definition: 432 =8 = Example: Given log255=12 , change the equation to Exponential form. Solution: Compare the equation to the definition and rewrite it. Definition: log = = Given: log255=12 Notice that =25, =5,and =12,respectively. Therefore, using the definition: log255= 12 = This instructional aid was prepared by the Tallahassee Community College Learning Commons. Solving Logarithm and Exponential equations Evaluate Logarithmic equations by using the definition of a logarithm to change the equation into a form that can then be solved.

3 Example: Given 3 1=7 , solve for . Solution: Step 1: Set up the equation and use the definition to change it. Definition: log = = Given 3 1=7 Notice 3 is the base or , and 7 is the given number. 3 1=7 log37= 1 Step 2: Now use the properties of Logarithms to solve. Recall the Change of Base Property: log =log log Apply it to log37. log37=log7log3 Step 3: Use the order of operations to finish solving for . 1= log7log3 = + Example: Given log6( +2)=3 , solve for . Solution: Step 1: Set up the equation and use the definition to change it.

4 Definition: log = = Given log6( +2)=3 Notice 6 is the base or , and 3 is the exponent or . log6( +2)=3 63= +2 Step 2: Now use the order of operations to solve. 63= +2 216= +2 214= = This instructional aid was prepared by the Tallahassee Community College Learning Commons. Expanding and Simplifying Logarithms To expand or simplify Logarithms , utilize the various properties of Logarithms in conjunction with the definition. Example: Given log3(9 2 2+1),expand the logarithm. Solution: Step 1: Expand the expression using the properties of Logarithms .

5 Recall the Logarithm Multiplication and Division Properties: log =log +log log ( )=log log Apply them to 9 2 and 2+1. Given log3(9 2 2+1): log39+log3 2 log3( 2+1) Step 2: Now simplify further using the properties of Logarithms and the definition. Recall the Logarithm for Powers Property: log = log Apply it to the 2andlog3( 2+1). log39+log3 2 log3( 2+1) log39+log3 2 log3( 2+1)12 log39+2log3 12log3( 2+1) By definition, log39=2 since 32=9, so our final answer becomes: + ( + ) Example: Write 3log2 log2 7log2 as a single logarithm.

6 Solution: To simplify the expression, work backwards with the Logarithmic properties. Step 1: Use the Logarithm for Powers Property where appropriate. Given: 3log2 log2 7log2 Notice that it can be applied to 3log2 and 7log2 . 3log2 log2 7log2 log2 3 log2 log2 7 Step 2: Simplify using the Logarithm Multiplication and Division Properties. Use the order of operations as a guide. log2 3 log2 log2 7 log2 3 (log2 +log2 7) log2 3 log2 7 This instructional aid was prepared by the Tallahassee Community College Learning Commons. Solving Expanded Logarithms Solving expanded Logarithms requires applying the definition of Logarithms and all the logarithm properties as needed.

7 Example: Given ln( 2)+ln( 3)=ln(2 +24),solve for . Solution: Note: ln( 2) is only valid if 2,ln( 3) is only valid if 3,and ln(2 +24) is only valid if the equation to be valid,all conditions must be met,so 3. Step 1: Simplify the left side of the equation using the multiplication and division properties of Logarithms . ln( 2)+ln( 3)=ln(2 +24) ln( 2)( 3)=ln(2 +24) ln( 2 5 +6)=ln(2 +24) Step 2: Use logarithm properties. Recall logarithm properties of bases: ln = and ln = ln( 2 5 +6)=ln(2 +24) Let both sides of the equation become the exponent of the base , and apply the property.

8 Ln( 2 5 +6)= ln(2 +24) 2 5 +6=2 +24 Step 3: Combine like terms to solve for . 2 5 +6=2 +24 2 7 18=0 ( 9)( +2)=0 =9, 2 Step 4: Check your answers. Recall that every logarithm must meet the conditions for the answer to be correct. For =9 ln((9) 2)+ln((9) 3)=ln(2(9)+24) ln(7)+ln(6)=ln(42) ln(7 6)=ln(42) This is valid! For = 2 Since 2 3, it does not meet all the conditions, and is not valid. Therefore: = Practice Exercises: 1. Given log4( )+log4(6 )=2, Solve for x. 2. Expand log2( 2 1) completely. 3. Write the following as a single logarithm: 2log3 +4 8log3 Answers: 1.

9 = 2 2. log2 12log2( 1) 12log2( +1) 3. log381 3 8