Indices u0026amp; the Law of Indices

Indices are a useful way of more simply expressing large numbers. They also present us with many useful properties for manipulating them using what are called the Law of Indices.

Indices

The expression 2⁵ is defined as follows:

We call “2” the base and “5” the index.

Law of Indices

To manipulate expressions, we can consider using the Law of Indices. These laws only apply to expressions with the same base, for example, 3⁴ and 3² can be manipulated using the Law of Indices, but we cannot use the Law of Indices to manipulate the expressions 3⁵ and 5⁷ as their base differs (their bases are 3 and 5, respectively).

Rule 1

Ex-2⁰= 1, 9⁰= 1, 208⁰= 1 and so on.

Rule 2:

Example 2^-2: 1/4, 6^-2: 1/36, 11^-2: 1/121 and so on.

Rule 3: 

a^m × aⁿ = a^m+n

: (note: 5 = 5¹)

Rule 4:

a^m ÷ aⁿ = a^m-n

Example-

:

Rule 5:

(a^m)ⁿ =a^mn

Example-  (y²)⁶:

Rule 6

Example-

125^2/3:

Question 1

Solution

Question 2

Solution

Question-3

Solution

Qustion-4

Solution

Question 5

Solution

ALGEBRIC EQUATION

VARIABLE

• The unknown quantities used in any equation are known as variables .
• Generally, they are denoted by the last english alphabets x, y, z.etc.
• An equation is a statement of equality of algebraic expressions, which involve one or more unknown Quantities , called the variables .
• LINEAR EQUATION
• An equation in which the highest power of variables is one, is called linear equation.These equations are called linear because the graph of such Equations on x-y Cartesian plane is a straight line.
• Linear equation in one variable :A  linear equation which contains only one variable is called linear equation in one variable.The general form of such equations is ax+b = cWhere a, b and c are constants and a not equal to 0.
• SOLUTION : All the values of x that satisfies this equation are called its solutions.
• Applications of linear equations with one variable.
• STEPS INVOLVED IN SOLVING A LINEAR EQUATION WORD
• PROBLEM
• Read the problem carefully and note what is given and what is required and what is given.
  ●Denote the unknown by the variables as x, y, …….
  ● Translate the problem to the language of mathematics or mathematical statements.
  ● Form the linear equation in one variable using the conditions given in the problems.
  ● Solve the equation for the unknown.
  ● Verify to be sure whether the answer satisfies the conditions of the problem.
• Q1 .The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find    the numbers.   Solution:Let the number be x. Then the other number = x + 9
  Sum of two numbers = 25
  According to question, x + x + 9 = 25⇒ 2x + 9 = 25
  ⇒ 2x = 25 – 9 (transposing 9 to the R.H.S)
  ⇒ 2x = 16
  ⇒ 2x/2 = 16/2 (divide by 2 on both the sides)
  ⇒ x = 8, Other number is x + 9 = 8 + 9 = 17
• Therefore, the two numbers are 8 and 17.
• Q-2. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages.
• Solution:Let Ron’s present age be x.
  Then Aaron’s present age = x – 5After 4 years Ron’s age = x + 4, Aaron’s age x – 5 + 4.
  According to question;Ron will be twice as old as Aaron.Therefore, x + 4 = 2(x – 5 + 4)
  ⇒ x + 4 = 2(x – 1) ⇒ x + 4 = 2x – 2
  ⇒ x + 4 = 2x – 2 ⇒ x – 2x = -2 – 4 ⇒ -x = -6 ⇒ x = 6
  Therefore, Aaron’s present age = x – 5 = 6 – 5 = 1
• Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.
• Q3. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair.
• The table cost $ 40 more than the chair.
  Let us assume the cost of the chair to be x.
  Then the cost of the table = x + $ 40
  The cost of 3 chairs = 3 × x = 3x andthe cost of 2 tables =2(40 + x)
  Total cost of 2 tables and 3 chairs = $705
  Therefore, 2(40 + x) + 3x = 705=> 80 + 2x + 3x = 705,=> 80 + 5x = 705
  => 5x = 705 – 80 => 5x = 625/5
  => x = 125 and 40 + x = 40 + 125 = 165
• Therefore, the cost of each chair is $125 and that of each table is $165.