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 then we introduce the inverse function logarithm and define

(Read as log base of equals).

ecolebooks.com

 In general

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:

 Þ Þ Þ Þ Þ Þ

And in going from logarithmic form to exponent form:

 Þ Þ Þ Þ Þ Þ

Laws of logarithms

Product and Quotient Laws of Logarithms:

The Product Law

The Quotient Law

Example.

= 2

The Power Law of Logarithms:

Example.

2log 5 + 2log 2

= 2

Logarithm of a Root

or

Example.

PROOF OF PROPERTIES

 Property Proof Reason for Step 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 = nb. x = bm and y = b nc. xy = bm * bnd. xy = b m + ne. logb xy = m + nf logb xy = logb x + logb y  a. Setupb. Rewrite in exponent formc. Multiply togetherd. Product rule for exponentse. Rewrite in log formf. Substitution (quotient rule)logb = logb x – logb y a. Let logb x = m and logb y = nb. x = bm and y = b nc. = d. = e. logb = m – nf. logblogb x – logb y  a. Given: compact formb. Rewrite in exponent form c. Divide d. Quotient rule for exponents e. Rewrite in log form f. Substitution (power rule)logb xn = n logb x a. Let m = logb x so x = bmb. xn = bmnc. logb x n = mnd. logb xn = n logb x a. Setupb. Raise both sides to the nth powerc. Rewrite as logd. 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 = 0ii. logbiii. =b0 iv. 1 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)

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:

(the one becomes an exponent : )

Verify:

not possible

Solving equations using logs

Examples

(i) Solve the equation

The definition of logs says ifthen or

Hence (to 5 decimal places)

Check (to 5 decimal places)

In practice from we take logs to base 10 giving

(ii) Solve the equation

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

(iii) Solve the equation

Check very close!

Note you could combine terms, giving,

(iv) Solve the equation

 Take logs of both sidesExpand bracketsCollect termsFactorise the left hand side divide

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

Check and

Notice that you could combine the log-terms in

to give

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

(v) Solve the equation

 Take logs of both sidesExpand bracketsCollect terms Factorize left hand side simplify  divide

Check

LHS = (taking )

RHS = (taking )

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

subscriber

By

By

By

By