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Solution :

(i)`((2^5)^2 * 7^3)/(8^3*7)` <br>
`= (2^(5*2)*7^3)/((2^3)^3*7)` <br>
`= (2^10*7^3)/(2^9*7)` <br>
`= 2*7^2 = 98` <br>
(ii) `(25*5^2*t^8)/(10^3*t^4)` <br>
`= (5^2*5^2*t^8)/(2^3*5^3*t^4)` <br>
`= (5^(4-3)*t^(8-4))/2^3` <br>
`= (5t^4)/8` <br>
answers**Introduction**

**In general if n is a natural `a xx a xx a xx a xx ...xx a = a^n `n - times a^n is called the nth power of a and is a also read as a raised to the n**

**In is evident from the above discussion that `a xx a xx a xx b xx b ` is written as `a^3 b^2` (read as a cubed into b squared ) ` a xx a xx b xx b xx b xx b` is written `a^2 b^4 ` (read as a squared into b raised to the power4)**

**For any non - zero rational number a we define `(i) a^1 = a` (ii) `a^@=1` .**

**If following form the above example that `(-1)` an odd natural number `=-1 and (-1)` An even natural number `=1` i.e. `(-1)^n={1` if n is an even natural number and `-1` if n is odd natural number**

**If `a/b` is a rational number and n is a natural number then `(a/b)^n=a/b xx a/b xx a/b xx a/b xx.....xx a/b = (a xx a xx a xx a xx....xx a)/(b xx b xx b xx b xx.......xx b)=a^n/b^n`**

**Express the following numbers as product of powers of their prime factors : (i) `1000` (ii) `16000` (iii) `3600`**

**Mul tipl ying powers with the same best**

**First law if is any non -zero rational number and` m n ` are natural numbers then `a^m xx a^n = a^(m+n) `**

**Generalisation : if a is a non - zero rational number and `m,n,p ` are natural numbers then `a^m xx a^n xx a^p = a^(m+n+p)`**