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Specific Objectives

By the end of the topic the learner should be able to:

 

(a) Derive logarithmic relation from index form and vice versa;

(b) State the laws of logarithms;

(c) Use logarithmic laws to simplify logarithmic expressions and solvelogarithmic equations;

(d) Apply laws of logarithms for further computations.

Content

(a) Logarithmic notation (eg. an=b, log ab=n)

(b) The laws of logarithms: log (AB) = log A + log B, log(A^B) = log A -log B and Log A n = n x log A.

(c) Simplifications of logarithmic expressions

(d) Solution of logarithmic equations

(e) Further computation using logarithmic laws.

If Image From EcoleBooks.comthen we introduce the inverse function logarithm and define Image From EcoleBooks.com

(Read as log base Image From EcoleBooks.com of Image From EcoleBooks.comequalsImage From EcoleBooks.com).

ecolebooks.com

 

In general

Image From EcoleBooks.com

 

 

Where Û means “implies and is implied by” i.e. it works both ways!

 

Note this means that, going from exponent form to logarithmic form:

Image From EcoleBooks.comÞ

Image From EcoleBooks.com

Image From EcoleBooks.comÞ

Image From EcoleBooks.com

Image From EcoleBooks.comÞ

Image From EcoleBooks.com

Image From EcoleBooks.comÞ

Image From EcoleBooks.com

Image From EcoleBooks.comÞ

Image From EcoleBooks.com

Image From EcoleBooks.comÞ

Image From EcoleBooks.com

 

And in going from logarithmic form to exponent form:

Image From EcoleBooks.comÞ

Image From EcoleBooks.com

Image From EcoleBooks.comÞ

Image From EcoleBooks.com

Image From EcoleBooks.comÞ

Image From EcoleBooks.com

Image From EcoleBooks.comÞ

Image From EcoleBooks.com

Image From EcoleBooks.comÞ

Image From EcoleBooks.com

Image From EcoleBooks.comÞ

Image From EcoleBooks.com

 

Laws of logarithms

Product and Quotient Laws of Logarithms:

 

Image From EcoleBooks.com The Product Law

Image From EcoleBooks.com The Quotient Law

 

 

Example.

Image From EcoleBooks.com

 Image From EcoleBooks.com  

 Image From EcoleBooks.com

 = 2  

 

 
 

The Power Law of Logarithms:

 

Image From EcoleBooks.com

 

Example.

2log 5 + 2log 2

 Image From EcoleBooks.com  

 Image From EcoleBooks.com  

 Image From EcoleBooks.com  

 = 2

 

Logarithm of a Root

 

Image From EcoleBooks.com or Image From EcoleBooks.com

 

 

 

Example.

Image From EcoleBooks.comImage From EcoleBooks.com

PROOF OF PROPERTIES

Property

Proof

Reason for Step

  1. logb b = 1 and logb 1 = 0

b1 = b  and b0 = 1 

Definition of logarithms

2.(product rule)

logb xy = logb x + logb y

a. Let logb x = m and logb y = n

b. x = bm and y = b n

c. xy = bm * bn

d. xy = b m + n

e. logb xy = m + n

f logb xy = logb x + logb y 

a. Setup

b. Rewrite in exponent form

c. Multiply together

d. Product rule for exponents

e. Rewrite in log form

f. Substitution

  1. (quotient rule)

    logbImage From EcoleBooks.com = logb x – logb y

a. Let logb x = m and logb y = n

b. x = bm and y = b n

c. Image From EcoleBooks.com = Image From EcoleBooks.com

d. Image From EcoleBooks.com= Image From EcoleBooks.com

e. logb Image From EcoleBooks.com= m – n

f. logbImage From EcoleBooks.comlogb x – logb y 

a. Given: compact form

b. Rewrite in exponent form

 

c. Divide

 

d. Quotient rule for exponents

 

e. Rewrite in log form

 

f. Substitution

  1. (power rule)

    logb xn = n logb x


     

a. Let m = logb x so x = bm

b. xn = bmn

c. logb x n = mn

d. logb xn = n logb x

a. Setup

b. Raise both sides to the nth power

c. Rewrite as log

d. Substitute

5. Properties used to solve log equations:

 

a. if bx = by, then x = y

 

 

b. if logb x = logb y, then x = y

 

 

 

a. This follows directly from the properties for exponents.

 

b. i. logb x – logb y = 0

ii. logbImage From EcoleBooks.com

iii. Image From EcoleBooks.com=b0

iv. Image From EcoleBooks.com1 so x = y

 

 

 

 

 

 

b. i. Subtract from both sides

 

ii. Quotient rule

 

 

iii. Rewrite in exponent form

 

 

iv. b0 = 1

 

 

Solving exponential and logarithmic equations

By taking logarithms, and exponential equation can be converted to a linear equation and solved. We will use the process of taking logarithms of both sides.

 

Example.

a) Image From EcoleBooks.com  

 Image From EcoleBooks.com

 Image From EcoleBooks.com

 Image From EcoleBooks.com  x = 1.792  

Note;

A logarithmic expression is defined only for positive values of the argument. When we solve a logarithmic equation it is essential to verify that the solution(s) does not result in the logarithm of a negative number. Solutions that would result in the logarithm of a negative number are called extraneous, and are not valid solutions.

 

Example.

Solve for x:

 Image From EcoleBooks.comImage From EcoleBooks.com (the one becomes an exponent : Image From EcoleBooks.com)

 Image From EcoleBooks.com

 Image From EcoleBooks.com

 Image From EcoleBooks.com

 Image From EcoleBooks.comImage From EcoleBooks.com

 

Verify:

Image From EcoleBooks.com Image From EcoleBooks.comnot possible

Solving equations using logs

 

Examples

(i) Solve the equation Image From EcoleBooks.com

The definition of logs says ifImage From EcoleBooks.comthen Image From EcoleBooks.com or Image From EcoleBooks.com

 Hence Image From EcoleBooks.com (to 5 decimal places)

Check Image From EcoleBooks.com(to 5 decimal places)

In practice from Image From EcoleBooks.com we take logs to base 10 giving

Image From EcoleBooks.com

 

(ii) Solve the equation Image From EcoleBooks.com

Image From EcoleBooks.com

 

Check Image From EcoleBooks.com, Image From EcoleBooks.com, we want Image From EcoleBooks.com so the value of Image From EcoleBooks.com lies between 3 and 4 orImage From EcoleBooks.com which means Image From EcoleBooks.com lies between 1.5 and 2. This tells us that Image From EcoleBooks.com is roughly correct.

(iii) Solve the equation Image From EcoleBooks.com

Image From EcoleBooks.com

 Check Image From EcoleBooks.comvery close!

Note you could combine terms, giving,

 Image From EcoleBooks.com

 

 

(iv) Solve the equation Image From EcoleBooks.com

 

 

Take logs of both sides

Expand brackets

Collect terms

Factorise the left hand side

 

divide

Image From EcoleBooks.com

(Note you get the same answer by using the ln button on your calculator.)

 

Check Image From EcoleBooks.comand Image From EcoleBooks.com

 

Notice that you could combine the log-terms in

Image From EcoleBooks.comto give Image From EcoleBooks.com

It does not really simplify things here but, in some cases, it can.

(v) Solve the equation Image From EcoleBooks.com

 

 

Take logs of both sides

Expand brackets

Collect terms

 

Factorize left hand side

 

simplify

 

 

divide

Image From EcoleBooks.com

 

Check

LHS = Image From EcoleBooks.com (taking Image From EcoleBooks.com)

RHS = Image From EcoleBooks.com (taking Image From EcoleBooks.com)

The values of LHS and RHS are roughly the same. A more exact check could be made using a calculator.

 

 

 

 

 

 

 

 

Logarithmic equations and expressions

Consider the following equations

 

The value of x in each case is established as follows

 

Therefore

 

X =4

 

 

 

 

Example

Solve

Solution

Let = t. then = 2

Introducing logarithm to base 10 on both sides

 

 

 

 

 

 

Therefore

 

 

 

Example

 

Taking logs on both sides cannot help in getting the value of x, since cannot be combined into a single expression. However if we let then the equation becomes quadratic in y.

Solution

Thus, let …………….. (1)

Therefore

 

 

Substituting for y in equation (1);

Let or let

There is no real value of x for which hence

 

Example

Solve for x in

Solution

Let

Therefore

solve the quadratic equation using any method

 

 

 

 

 

Substituting for t in the equation (1).

 

= x

 

Note;

 

 

End of topic  

Did you understand everything?

If not ask a teacher, friends or anybody and make sure you understand before going to sleep!

 

Past KCSE Questions on the topic.

 

1.  Solve for ( – ½ = 3/2

2.  Find the values of x which satisfy the equation 52x – 6 (5x) + 5 =0

 

3.  Solve the equation

 Log (x + 24) – 2 log 3 = log (9-2x)

4. Find the value of x in the following equation 49(x+1) + 7(2x) = 350

5. Find x if 3 log 5 + log x2 = log 1/125

6. Without using logarithm tables, find the value of x in the equation

Log x3 + log 5x = 5 log2 – log 2 5

7. Given that P = 3y express the questions 32y -1) + 2 x 3(y-1) = 1 in terms of P

8.  Hence or otherwise find the value of y in the equation: 3(2y-1) + 2 x 3(y-1) =1




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EcoleBooks | Mathematics Form 1-4 : CHAPTER FOURTY FIVE - FURTHER LOGARITHMS

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