## Exponents

Published in: Mathematics
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• ### Nilanjana S

• Kolkata
• 16 Years of Experience
• Qualification: M.Sc
• Teaches: Mathematics
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This deals with the very basics of the exponents, the laws of exponents and their validity.

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Ùˆsweuodx
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1. 2. 3. Today we will learn what do we mean by the term exponent or index The need for exponents Laws of indices and their verification
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We all know how difficult it is to write the distance between the earth and the sun and other planetary distances which are very huge. The distance from the earth to the sun is 149,597,870,700 meters. On the other end, there are very small distances like the distances between the sub atomic particles which are equally hard to express. In these situations, expressing a number in its exponential form is very useful. We have learnt that b+b+b+b+b = 5b b5 which is read as b raised to the power of 5 . Here, b is the base and 5 is the Similarly bxbxbxbxb = exponent or index. b2 is b raised to the power of 2 or b squared. bxb = bxbxb = b3 is b raised to the power of 3 or b cubed. So instead of writing 1000000000000, we can write 1012 since 1000000000000 = = exponent. In general axaxaxax m times = am 1012. Here 10 is the base and 12 is the Here 'a' is the base and 'm' is the exponent or power and is read as 'a' raised to the power of 'm'.
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Negative Power x-i is x raised to the power of -1. This is equal to l/x and is called the reciprocal of x. So, expressing it in equation form we have l/x= xl 1/x2 = x-2 --- reciprocal of Similarly , 1/x3 = x-3 --- reciprocal of x 3 and so on. In general, l/xm Laws Of Indices Let a be any number not equal to 0. m and n are any two +ve integers 1. 2. 3. 4. 5. 6. i) am xan = am+n ii) am xan x ap x. m+n+p+.... am = am = amn ii) am ) n amnp (a x = am x b If 'a' is a positive number other than 1 and if am = an then m = n
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To find the meaning of ao using law no.2 we know that if we divide something by itself , the quotient is always 1. So we can write am m am-m -------------(2) But by law 2 we know that am Combining equations (1) and (2) we have ao 1 Solved Examples Ex 1. Find the value of 22 x 23 Soln 1 .Using Law 1 we can write that 22 x 23 = 22+3 = 25 = x 2 x 2 =32 Check We know that 22 = 2 x 2=4 Also But 4 x 8 =32 So our first answer is correct. Ex 2. Find the value of 32 x 31 x 30 Soln 2. Using Law 1 we can write 32 x 31 x 30 Ex 3. Find the value of 24+ 22 Soln 3. Using Law 2 we can write 24 + 22 = 24-2 Check 32+1+0 = 33 But 16 +4=4 So our first answer is correct.
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Ex 4. Find the value of ( 22 Soln 4. Using Law 3 we find that (22 = 22x3 Again 64 so ( 22 = 64 Check 22=2 x = 62 = 26 43 = 4 x 4 x 4 = 64 So our first answer is correct. Ex 5. Find the value of (2 x Soln 5. Using Law 4 we find that (2 x = 22 x 32 But 22=4 and 32=9 so 36 Check But 62 = 6 x 36 So our first answer is correct. Ex 6. Find the value of (6 + Soln 6. Using Law 5 we find that (6 22 But 62=36 and 22=4 so 36 +4=9 Check 6+2=3 And 9 So our first answer is correct.

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