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Ncert Solutions
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What is exponent and Base?
This can be explained with the below example
We know
2
2 = 2×2
3
3 = 3×3×3
So it is known how we interact with positive exponent
Negative exponents
Here we will be looking how to interact with these type of expression
2-10
11-1
Here we can see that exponents are negative
Now negative exponents can be converted into positive exponents like this
2
-10 = 1/2
10
11
-1= 1/11
So
we can say that for any non-zero integer
a,
a
-m = 1/a
m
where
m is a positive integer.
am is the multiplicative inverse of
a-m
Example
a) 2-3
b) 3-2
Solution
As a-m = 1/am
a) 2-3 = ½3 = 1/8
b) 3-2 = 1/32 =1/9
How to express the decimal number in exponent form
We already know that any non -decimal number can be expressed in exponent form like below
1215 =1000+ 200 +10 +5 = 1 × 10
3 + 2 × 10
2 + 2 × 10
1 + 5 × 10
0
So these are all positive exponents
Decimal number can be expressed as exponents also
1215.15 =1000+ 2-00 +10 +5 +.10+.05= 1 × 10
3 + 2 × 10
2 + 2 × 10
1 + 5 × 10
0 +1/10 + 5/100
=1 × 10
3 + 2 × 10
2 + 2 × 10
1 + 5 × 10
0 +10
-1 + 5 ×10
-2
So this has both the negative and positive exponents.
Laws of Exponents
Here are the laws of exponents when
a and
b are non-zero integers and
m,
n are any integers.
a
-m = 1/a
m
a
m / a
n = a
m-n
(a
m )
n = a
mn
a
m x b
m = (ab)
m
a
m / b
m = (a/b)
m
a
0 =1
(a/b)
-m =(b/a)
m
(1)
n = 1 for infinitely many
n.
(-1)
p =1 for any even integer p
These laws can be used to solve the exponents problems
Example:
(-3)3 × (2/3)3
Solution
(-3)3 × (2/3)3
=(-1)3 × (3)3 × (2)3 × (3)-3 as am x bm = (ab)m and am / bm = (a/b)m
= (-1)3 × (2)3
=-8
Important Note
an = 1 only if n = 0. This will work for any a except a = 1 or a = –1. For a = 1, 11 = 12 =13 = 1 or (1)n = 1 for infinitely many n.
For a = –1,(–1)0 = (–1)2 = (–1)4 = (–1)-2 = ... = 1 or (–1)p = 1 for any even integer p.
Watch this tutorial for more explanation About exponents and power
Use of Exponents to Express Small Numbers in Standard Form
In previous classes we have learnt how to convert large number into exponents form. We can express small number in the standard exponent form also
Let’s take an example to explain
.00001
Step 1: Count of Number of places after decimal
Here it is 5
Step 2: On the basis of counting in step 1,the number can be written as
.00001 = 1/100000=1/105 = 1X 10-5
as am = 1/a-m
Basically the decimal has been moved to the place after digit
Another example would be
.00019
Step 1: In these cases, we count of Number of places after decimal to the first digit only
Here it is 4
Step 2: On the basis of counting in step 1,the number can be written as
.0001 = 1.9/10000=1.9/104 = 1.9X 10-4
as am = 1/a-m
Watch this tutorial for more explanation About Use of Exponents to Express Small Numbers in Standard Form
Practice Questions
a) .00156
b) .005402
c) .1234
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Class 8 Maths
Class 8 Science
