Chapter-9 (SURDS AND INDICES)

9. SURDS AND INDICES

I IMPORTANT FACTS AND FORMULAE I

1. LAWS OF INDICES:

(i)                 a^m x aⁿ = a^{m + n}

(ii)    a^m_­ / aⁿ = a^m-n

(iii)             (a^m)ⁿ = a^mn

(iv)             (ab)ⁿ = aⁿbⁿ

(v)     ( a/ b )ⁿ = ( aⁿ / bⁿ )

(vi)             a⁰ = 1

                                                                                                        

2. SURDS: Let a be a rational number and n be a positive integer such that a^1/n = ⁿsqrt(a)

   is irrational. Then ⁿsqrt(a)  is called a surd of order n.

3. LAWS OF SURDS:

(i)  ⁿ√a = a^1/2

(ii) ⁿ √ab = ⁿ √a * ⁿ √b

(iii) ⁿ √a/b = ⁿ √a  /  ⁿ √b

(iv) (ⁿ √a)ⁿ = a

(v) ^m√(ⁿ√(a)) = ^mn√(a)

(vi) (ⁿ√a)^m = ⁿ√a^m

I SOLVED EXAMPLES

Ex. 1. Simplify : (i) (27)^2/3       (ii) (1024)^-4/5    (iii)( 8 / 125 )^-4/3

Sol .     (i) (27)^2/3 = (3³)^2/3 = 3^{( 3 * ( 2/ 3))} = 3² = 9

            (ii) (1024)^-4/5 = (4⁵)^-4/5 = 4 ^{{ 5 * ( (-4) / 5 )}} = 4^-4 = 1 / 4⁴ = 1 / 256

            (iii) ( 8 / 125 )^-4/3 = {(2/5)³}^-4/3 = (2/5)^{{ 3 * ( -4/3)}} = ( 2 / 5 )^-4 = ( 5 / 2 )⁴  = 5⁴ / 2⁴ = 625 / 16

           

Ex. 2. Evaluate: (i) (.00032)^3/5            (ii)l (256)^0.16 x (16)^0.18.

Sol.      (i) (0.00032)^3/5 = ( 32 / 100000 )^3/5. = (2⁵ / 10⁵)^3/5  =  {( 2 / 10 )⁵}^3/5 = ( 1 / 5 )^{(5 * 3 / 5)} =  (1/5)³  =   1 / 125

            (ii) (256)^{0. 16} * (16)^{0. 18} = {(16)²}^{0. 16} * (16)^{0. 18} = (16)^{(2 * 0. 16)} * (16)^{0. 18}

                                                =(16)^0.32 * (16)^0.18  = (16)^(0.32+0.18)  = (16)^0.5 = (16)^1/2 = 4.

  196

Ex. 3. What is the quotient when (x^-1 - 1) is divided by (x - 1) ?

Sol.     x^-1 -1 = (1/x)-1 = _1 -x *   1      = -1

            x - 1       x - 1           x     (x - 1)      x

Hence, the required quotient is    -1/x

Ex. 4. If 2^{x - 1} + 2^{x + 1} = 1280, then find the value of  x.

Sol.      2^{x - 1} + 2^{X+ 1} = 1280  ó 2^x-1 (1 +2²) = 1280

                                             ó 2^x-1  =   1280 / 5 = 256 =  2⁸   ó x -1 = 8 ó x  =  9.

                                                            

Hence, x = 9.

Ex. 5. Find the value of [ 5 ( 8^1/3 + 27^1/3)³]^{1/ 4}

Sol.      [ 5 ( 8^1/3 + 27^1/3)³]^{1/ 4}  =  [ 5 { (2³)^1/3 + (3³)^1/3}³]^{1/ 4} =   [ 5 { (2^{3 * 1/3})^1/3 + (3^{3 *1/3} )^1/3}³]^{1/ 4}

                                                = {5(2+3)³}^1/4 = (5 * 5³)^{1/ 4} =5^{(4 * 1/ 4)}  = 5¹ = 5.

Ex. 6. Find the Value of {(16)^3/2 + (16)^-3/2}

Sol.     [(16)^3/2 +(16)^-3/2 = (4²)^3/2 +(4²)^-3/2 = 4^{(2 * 3/2)} + 4^{{ 2* (-3/2)}}

                                 = 4³ + 4^-3 = 4³ + (1/4³) = ( 64 + ( 1/64)) = 4097/64.

Ex. 7. If (1/5)^3y = 0.008, then find the value of(0.25)^y.

Sol. (1/5)^3y = 0.008 =  8/1000 =  1/125 = (1/5)³ ó 3y = 3 ó Y = 1.

          \ (0.25)^y = (0.25)¹ = 0.25.

Ex. 8. Find the value of     (243)^n/5 ´ 3^{2n + 1}

                                             9ⁿ ´ 3^{n -1} .

Sol. (243)^n/5 x3^2n+l   =   3 ^{(5 * n/5)} ´ 3^2n+l _ = 3ⁿ  ´3²ⁿ⁺¹

       (3²)ⁿ ´ 3^{n - 1}          3²ⁿ ´ 3^{n - 1}                 32n ´ 3^n-l

                                    = 3^{n + (2n + 1)}   =   3⁽³ⁿ⁺¹⁾   =  3^{(3n+l)-(3n-l)} = 3² = 9.

                                          3^2n+n-1           3^(3n-1)

Ex. 9. Find the value Of (2^1/4-1)(2^3/4+2^1/2+2^1/4+1)

Sol.

            Putting 2^1/4 = x, we get :

            (2^1/4-1) (2^3/4+2^1/2+2^1/4+1)=(x-1)(x³+x²+x+1) , where x = 2^1/4

                                                     =(x-1)[x²(x+1)+(x+1)]

                                                     =(x-1)(x+1)(x²+1) = (x²-1)(x²+1)

                                                     =(x⁴-1) = [(2^1/4)⁴-1] = [2^(1/4*⁴⁾ –1] = (2-1) = 1.

           

Ex. 10. Find the value of   6^2/3 ´  ³√6⁷

                                                ³√6⁶

Sol.       6^2/3 ´  ³√6⁷   =  6^2/3  ´ (6⁷)^1/3   =  6^2/3  ´  6^{(7 * 1/3)}  =   6^2/3  ´  6^(7/3)

                  ³√6⁶                 (6⁶)^1/3                  6^{(6 * 1/3)}                    6²

                                    =6^2/3 ´ 6^((7/3)-2) = 6^2/3 ´ 6^1/3  = 6¹ = 6.

Ex. 11. If x= y^a, y=z^b and z=x^c,then find the value of abc.

Sol.    z¹= x^c =(y^a)^c       [since x= y^a]

             =y^(ac) = (z^b)^ac   [since y=z^b]

             =z^b(ac)= z^abc

\           abc = 1.          

 = 24

                                  Ex. 12. Simplify [(x^a / x^b)^(a²+b²+ab)] * [(x^b / x^c )^ b²+c²+bc)] * [(x^c/x^a)^(c²+a²+ca)]

Sol.

 Given Expression

= [{x^{(o - b)}}^(a² + b² + ob)].['(x^{(b - c)}}^ (b² + c² + bc)].['(x^{(c - a)}}^(c² + a² + ca])

= [x^{(a - b)(a2 + b2 + ab)} . x^{(b - c) (b2 +c2+ bc)}.x^{(c- a) (c2 + a2 + ca)}]

= [x^(a³-b³)].[x^(b³-e³)].[x^(c³-a³)] = x^(a³-b³+b³-c³+c³-a³) = x⁰ = 1.

Ex. 13. Which is larger √2 or ³√3 ?

Sol. Given surds are of order 2 and 3. Their L.C.M. is 6. Changing each to a surd of order 6, we get:

                             √2 = 2^1/2 = 2^{((1/2)*(3/2))} =2^3/6 =  8^1/6 = ⁶√8

                             ³√3= 3^1/3 = 3^{((1/3)*(2/2))} = 3^2/6 = (3²)^1/6 = (9)^1/6 = ⁶√9.

Clearly, ⁶√9 > ⁶√8 and hence ³√3  > √2.

Ex. 14. Find the largest from among 4√6, √2 and ³√4.

Sol. Given surds are of order 4, 2 and 3 respectively. Their L.C,M, is 12, Changing each to a surd of order 12, we get:

⁴√6 = 6^1/4 = 6^{((1/4)*(3/3))} = 6^3/12 = (6³)^1/12  = (216)^1/12.

√2 = 2^1/2 = 2^{((1/2)*(6/6))} = 2^6/12 = (2⁶)^1/12  = (64)^1/12.

³√4 = 4^1/3 = 4^{((1/3)*(4/4))}  =  4^4/12  = (4⁴)^1/12 = (256)^1/12.

Clearly, (256)^1/12  >  (216)^1/12  >  (64)^1/12

Largest one is (256)^1/12.  i.e. ³√4 .

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