C) solving equations

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

C) solving equations

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C) solving equations