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Objective 
Solving equations involving indices 
and logarithms
Warm up 
 Q1: Express it in index form. 
log x = 2 a 
Solution: 
 Q2: Express it in logarithmic form 
Solution : 
a2 = x 
3 
4 
27 = 81 
log 27 3 81 = 
4
When working with logarithms, 
if ever you get “stuck”, try 
rewriting the problem in 
exponential form. 
Conversely, when working 
with exponential expressions, 
if ever you get “stuck”, try 
rewriting the problem 
in logarithmic form.
Let’s see if this simple 
rule 
can help us solve some 
of the following 
problems.
Continuous assessment 
Example 1: 
Solve for x: log6 x = 2 
Solution: 
Let’s rewrite the problem 
in exponential form. 
62 = x 
We’re finished !
Example 2 
Solve for y: log 1 
Solution: 
5 
Rewrite the problem in 
exponential form. 
5y = 
1 
25 
Since 
æ  
1 
çè   
= 5- 2 25 
ö  
÷ø   
5y = 5- 2 
y = -2 
25 
= y
Example 3 
Evaluate log3 27. 
Solution: 
Try setting this up like this: 
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: 
log7 72 = y First, we write the problem with a variable. 
Now take it out of the logarithmic form 
and write it in exponential form. 
7y = 72 
y = 2
Example 5 
Evaluate: 4log4 16 
Solution: 
4log4 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
Finally, we want to take a look at 
the Property of Equality for 
Logarithmic Functions. 
Suppose b > 0 and b ¹ 1. 
Then logb x1 = logb x2 if and only if x1 = x2 
Basically, with logarithmic functions, 
if the bases match on both sides of the equal 
sign , then simply set the arguments equal.
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 
Factor 
(x - 7)(x + 2) = 0 
(x - 7) = 0 or (x + 2) = 0 
x = 7 or x = -2 continued on the next page
Example 2 
continued 
Solve: log8 (x2 -14) = log8 (5x) 
Solution: 
x = 7 or x = -2 
It appears that we have 2 solutions here. 
If we take a closer look at the definition of 
a logarithm however, we will see that not 
only must we use positive bases, but also we 
see that the arguments must be positive as 
well. Therefore -2 is not a solution. 
Let’s end this lesson by taking a closer look 
at this.
Our final concern then is to 
determine why logarithms like 
the one below are undefined. 
log2 (-8) 
Can anyone give us 
an explanation ?
2 log (-8) = undefined WHY? 
One easy explanation is to simply rewrite 
this logarithm in exponential form. 
We’ll then see why a negative value is not 
permitted. 
First, we write the problem with a variable. 
2 log (-8) = y 
2y = - 8 Now take it out of the logarithmic form 
and write it in exponential form. 
What power of 2 would gives us -8 ? 
23 = 8 and 2- 3 = 
1 
8 
Hence expressions of this type are undefined.
That concludes our introduction 
to logarithms. In the lessons to 
follow we will learn some important 
properties of logarithms. 
One of these properties will give 
us a very important tool 
which 
we need to solve exponential 
equations. Until then let’s 
practice with the basic themes 
of this lesson.
Final assessment 
 Solve for x 
3 2x – 10 x 3x + 9 = 0
Home work 
 Solve for x 
= 
- 
x 
)7 1 
= 
a 
) log 4 2 
+ 
x 
x x 
1 
- ´ - = 
´ = 
b 
)2 2 10 
x + 
x 
2 1 
)3 26 3 9 0 
c 
d

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C) solving equations

  • 1. Objective Solving equations involving indices and logarithms
  • 2. Warm up  Q1: Express it in index form. log x = 2 a Solution:  Q2: Express it in logarithmic form Solution : a2 = x 3 4 27 = 81 log 27 3 81 = 4
  • 3. When working with logarithms, if ever you get “stuck”, try rewriting the problem in exponential form. Conversely, when working with exponential expressions, if ever you get “stuck”, try rewriting the problem in logarithmic form.
  • 4. Let’s see if this simple rule can help us solve some of the following problems.
  • 5. Continuous assessment Example 1: Solve for x: log6 x = 2 Solution: Let’s rewrite the problem in exponential form. 62 = x We’re finished !
  • 6. Example 2 Solve for y: log 1 Solution: 5 Rewrite the problem in exponential form. 5y = 1 25 Since æ 1 çè = 5- 2 25 ö ÷ø 5y = 5- 2 y = -2 25 = y
  • 7. Example 3 Evaluate log3 27. Solution: Try setting this up like this: log3 27 = y Now rewrite in exponential form. 3y = 27 3y = 33 y = 3
  • 8. 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.
  • 9. Example 4 Evaluate: log7 72 Solution: log7 72 = y First, we write the problem with a variable. Now take it out of the logarithmic form and write it in exponential form. 7y = 72 y = 2
  • 10. Example 5 Evaluate: 4log4 16 Solution: 4log4 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
  • 11. Finally, we want to take a look at the Property of Equality for Logarithmic Functions. Suppose b > 0 and b ¹ 1. Then logb x1 = logb x2 if and only if x1 = x2 Basically, with logarithmic functions, if the bases match on both sides of the equal sign , then simply set the arguments equal.
  • 12. 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
  • 13. 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 Factor (x - 7)(x + 2) = 0 (x - 7) = 0 or (x + 2) = 0 x = 7 or x = -2 continued on the next page
  • 14. Example 2 continued Solve: log8 (x2 -14) = log8 (5x) Solution: x = 7 or x = -2 It appears that we have 2 solutions here. If we take a closer look at the definition of a logarithm however, we will see that not only must we use positive bases, but also we see that the arguments must be positive as well. Therefore -2 is not a solution. Let’s end this lesson by taking a closer look at this.
  • 15. Our final concern then is to determine why logarithms like the one below are undefined. log2 (-8) Can anyone give us an explanation ?
  • 16. 2 log (-8) = undefined WHY? One easy explanation is to simply rewrite this logarithm in exponential form. We’ll then see why a negative value is not permitted. First, we write the problem with a variable. 2 log (-8) = y 2y = - 8 Now take it out of the logarithmic form and write it in exponential form. What power of 2 would gives us -8 ? 23 = 8 and 2- 3 = 1 8 Hence expressions of this type are undefined.
  • 17. That concludes our introduction to logarithms. In the lessons to follow we will learn some important properties of logarithms. One of these properties will give us a very important tool which we need to solve exponential equations. Until then let’s practice with the basic themes of this lesson.
  • 18. Final assessment  Solve for x 3 2x – 10 x 3x + 9 = 0
  • 19. Home work  Solve for x = - x )7 1 = a ) log 4 2 + x x x 1 - ´ - = ´ = b )2 2 10 x + x 2 1 )3 26 3 9 0 c d