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LOGARITHMIC FUNCTION,
LOGARITHMIC EQUATION AND
LOGARITHMIC INEQUALITY
 FELINA E. VICTORIA
 MAED-MATH

Example 1:
Solution: log2 8  3
We read this as: ”the log
base 2 of 8 is equal to 3”.
3
Write 2 8 in logarithmic form.
Example 1a:
Write 42
16 in logarithmic form.
Solution: log4 16  2
Read as: “the log
base 4 of 16 is
equal to 2”.
Example 1b:
Solution:
Write 2
 3

1
8
in logarithmic form.
log2
1
8
  3
1
Read as: "the log base 2 of is equal to -3".
8
Okay, so now it’s time for
you to try some on your
own.
1. Write 72
 49 in logarithmic form.
7Solution: log 49 2
log5 1 0Solution:
2. Write 50
 1 in logarithmic form.
3. Write 10
 2

1
100
in logarithmic form.
Solution: log10
1
100
  2
Solution: log16 4 
1
2
4. Finally, write 16
1
2
 4
in logarithmic form.
Example 1:
Write log3 81 4 in exp onential form
Solution: 34
 81
Example 2:
Write log2
1
8
  3 in exp onential form.
Solution: 2
 3

1
8
Okay, now you try these next
three.
1. Write log10 100  2 in exp onential form.
3. Write log27 3 
1
3
in exp onential form.
2. Write log5
1
125
  3 in exp onential form.
1. Write log10 100  2 in exp onential form.
Solution: 102
100
2. Write log5
1
125
  3 in exp onential form.
Solution: 3 1
5
125


3. Write log27 3 
1
3
in exp onential form.
Solution: 27
1
3
 3
Solution:
Let’s rewrite the problem
in exponential form.
62
 x
We’re finished !
6Solve for x: log 2x 
Example 1
Solution:
5
y

1
25
Rewrite the problem in
exponential form.
Since
1
25
 5 2





5y
 5 2
y  2
5
1
Solve for y: log
25
y
Example 2
Example 3
Evaluate log3 27.
Try setting this up like this:
Solution:
log3 27  y Now rewrite in exponential form.
3y
 27
3y
 33
y  3
These next two problems
tend to be some of the
trickiest to evaluate.
Actually, they are merely
identities and
the use of our simple
rule
will show this.
Example 4
Evaluate: log7 72
Solution:
Now take it out of the logarithmic form
and write it in exponential form.
log7 72
 y First, we write the problem with a variable.
7y
 72
y  2
Example 5
Evaluate: 4log 4 16
Solution:
4
log 4 16
 y First, we write the problem with a variable.
log4 y  log4 16 Now take it out of the exponential form
and write it in logarithmic form.
Just like 23
 8 converts to log2 8  3
y  16
Example 1
Solve: log3 (4x 10)  log3 (x 1)
Solution:
Since the bases are both ‘3’ we simply set
the arguments equal.
4x10 x1
3x101
3x   9
x   3
Example 2
Solve: log8 (x2
14)  log8 (5x)
Solution:
Since the bases are both ‘8’ we simply set the arguments equal.
x2
14  5x
x2
 5x 14  0
(x  7)(x  2) 0
Factor
(x  7) 0 or (x  2) 0
x  7 or x  2 continued on the next page

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Logarithmic function, equation and inequality

  • 1. LOGARITHMIC FUNCTION, LOGARITHMIC EQUATION AND LOGARITHMIC INEQUALITY  FELINA E. VICTORIA  MAED-MATH
  • 2.
  • 3. Example 1: Solution: log2 8  3 We read this as: ”the log base 2 of 8 is equal to 3”. 3 Write 2 8 in logarithmic form.
  • 4. Example 1a: Write 42 16 in logarithmic form. Solution: log4 16  2 Read as: “the log base 4 of 16 is equal to 2”.
  • 5. Example 1b: Solution: Write 2  3  1 8 in logarithmic form. log2 1 8   3 1 Read as: "the log base 2 of is equal to -3". 8
  • 6. Okay, so now it’s time for you to try some on your own. 1. Write 72  49 in logarithmic form. 7Solution: log 49 2
  • 7. log5 1 0Solution: 2. Write 50  1 in logarithmic form.
  • 8. 3. Write 10  2  1 100 in logarithmic form. Solution: log10 1 100   2
  • 9. Solution: log16 4  1 2 4. Finally, write 16 1 2  4 in logarithmic form.
  • 10. Example 1: Write log3 81 4 in exp onential form Solution: 34  81
  • 11. Example 2: Write log2 1 8   3 in exp onential form. Solution: 2  3  1 8
  • 12. Okay, now you try these next three. 1. Write log10 100  2 in exp onential form. 3. Write log27 3  1 3 in exp onential form. 2. Write log5 1 125   3 in exp onential form.
  • 13. 1. Write log10 100  2 in exp onential form. Solution: 102 100
  • 14. 2. Write log5 1 125   3 in exp onential form. Solution: 3 1 5 125  
  • 15. 3. Write log27 3  1 3 in exp onential form. Solution: 27 1 3  3
  • 16. Solution: Let’s rewrite the problem in exponential form. 62  x We’re finished ! 6Solve for x: log 2x  Example 1
  • 17. Solution: 5 y  1 25 Rewrite the problem in exponential form. Since 1 25  5 2      5y  5 2 y  2 5 1 Solve for y: log 25 y Example 2
  • 18. Example 3 Evaluate log3 27. Try setting this up like this: Solution: log3 27  y Now rewrite in exponential form. 3y  27 3y  33 y  3
  • 19. These next two problems tend to be some of the trickiest to evaluate. Actually, they are merely identities and the use of our simple rule will show this.
  • 20. Example 4 Evaluate: log7 72 Solution: Now take it out of the logarithmic form and write it in exponential form. log7 72  y First, we write the problem with a variable. 7y  72 y  2
  • 21. Example 5 Evaluate: 4log 4 16 Solution: 4 log 4 16  y First, we write the problem with a variable. log4 y  log4 16 Now take it out of the exponential form and write it in logarithmic form. Just like 23  8 converts to log2 8  3 y  16
  • 22. Example 1 Solve: log3 (4x 10)  log3 (x 1) Solution: Since the bases are both ‘3’ we simply set the arguments equal. 4x10 x1 3x101 3x   9 x   3
  • 23. Example 2 Solve: log8 (x2 14)  log8 (5x) Solution: Since the bases are both ‘8’ we simply set the arguments equal. x2 14  5x x2  5x 14  0 (x  7)(x  2) 0 Factor (x  7) 0 or (x  2) 0 x  7 or x  2 continued on the next page