SURDS AND INDICES

Surds

Let a be a rational number and n be a positive integer such that 

a1/n =  n√a

Then, n√a is called a surd of order n. 

Laws of Surds

Law 1 : 

n√a  =  a1/n

Law 2 : 

n√(ab)  =  n√a x n√b

Law 3 : 

n√(a/b)  =  n√a / n√b

Law 4 : 

(n√a)n  =  a

Law 5 : 

m√(n√a)  =  mna

Law 6 : 

(n√a)m  =  n√am

Laws of Indices

Law 1 : 

xm ⋅ xn  =  xm+n

Law 2 : 

xm ÷ xn  =  xm-n

Law 3 : 

(xm)n  =  xmn

Law 4 : 

(xy)m  =  xm ⋅ ym

Law 5 : 

(x / y)m  =  xm / ym

Law 6 : 

x-m  =  1 / xm

Law 7 : 

x0  =  1

Law 8 : 

x1  =  x

Law 9 : 

xm/n  =  y -----> x  =  yn/m

Law 10 : 

(x / y)-m  =  (y / x)m

Law 11 : 

ax  =  ay -----> x  =  y

Law 12 : 

xa  =  ya -----> x  =  y

Solved Problems

Problem 1 : 

If x1/p  =  y1/q  =  z1/r and xyz  =  1, then find  the value of
(p + q + r).  

Solution : 

Let  x1/p  =  y1/q  =  z1/r  =  k.

Then,

x1/p  =  k -----> x  =  kp

y1/q  =  k -----> y  =  kpq

z1/r  =  k -----> z  =  kr

Given : xyz  =  1

Then,

xyz  =  1

k k kr  =  1

kp + q + r  =  1 -----(1)

We know that a0  =  1.

So,

k0 = 1

In (1),  substitute 1  =  k0.

(1) -----> kp + q + r  =  k0

Using law 11 of indices, we get

p + q + r  =  0

Problem 2 : 

Simplify : 

[1-{1-(1-x2)-1}-1]-1/2

Solution : 

[1-{1-(1-x2)-1}-1]-1/2  :

=  [1-{1-1/(1-x2)}-1]-1/2

=  [1-{(1-x2-1)/(1-x2)}-1]-1/2

=  [1-{-x2/(1-x2)}-1]-1/2

=  [1-{x2/(x2-1)}-1]-1/2

=  [1-(x2-1)/x2]-1/2

=  [{x2-(x2-1)}/x2]-1/2

=  [(x2-x2+1)/x2]-1/2

=  [1/x2]-1/2

=  [x2]1/2

=  x

Problem 3 : 

Using (a - b)3 = a3 - b3 - 3ab(a-b), if x  =  p1/3 - p-1/3, find the value of 

x3 + 3x

Solution : 

Given : x = p1/3 - p-1/3

Take power 3 on both the sides.

x3  =  (p1/3 - p-1/3)3

Using (a - b)3  =  a3 - b3 - 3ab(a - b).

x3  =  (p1/3)3 - (p-1/3)3 - 3p1/3.p-1/3(p1/3-p-1/3)

x =  p - p-1 - 3p1/3 - 1/3(x)

x =  p - 1/p - 3p0x

x =  p - 1/p - 3(1)x

x3  =  p - 1/p - 3x

x3 + 3x  =  p - 1/p

Problem 4 : 

Simplify : 

Solution : 

Problem 5 : 

If x  =  31/3 + 3-1/3, find the value of 

3x3 - 9x

Solution : 

Given : x = 31/3 + 3-1/3

Take power 3 on both the sides.

x3  =  (31/3 + 3-1/3)3

Using (a + b)3  =  a3 + b3 + 3ab(a + b).

x3  =  (31/3)3 + (3-1/3)3 + 3 ⋅ 31/3 ⋅ 3-1/3(31/3 + 3-1/3)

x =  3 + 3-1 + 3 ⋅ 31/3 - 1/3 x

x =  3 + 1/3 + 3 ⋅ 3⋅ x

x =  3 + 1/3 + 3(1)x

x3  =  3 + 1/3 + 3x

x3 - 3x  =  3 + 1/3

Multiply each side by 3.

3(x3 - 3x)  =  3(3 + 1/3)

3x3 - 9x  =  9 + 1

3x3 - 9x  =  10

Problem 6 :

If ax = b, by = c and  cz = a, then find the value of xyz. 

Solution : 

Let

ax  =  b -----(1)

by  =  c -----(2)

cz  =  a -----(3)

Substitute a  =  cin (1).

(1)-----> (cz)x  =  b

czx  =  b

Substitute c  =  by.

(by)zx  =  b

bxyz  =  b

bxyz  =  b1

xyz  =  1

Problem 7 :

If  2x  =  3y  =  6-z, then find the value of

1/x  +  1/y  +  1/z  

Solution : 

Let 2x  =  3y  =  6-z  =  k.

Then, 

2x  =  k ----->  2  =  k1/x

3y  =  k -----> 3  =  k1/y

6-z  =  k -----> 6  =  k-1/z

And also, 

6  =  k-1/z 

(2 x 3)  =  k-1/z

In (1), substitute 2  =  k1/x, 3  =  k1/y.

k1/x ⋅ k1/y  =  k-1/z

k1/x + 1/y  =  k-1/z

Using law 11 of indices, we get

1/x + 1/y  =  -1/z

1/x + 1/y + 1/z  =  0

Problem 8 :

If (√9)-7 ⋅ (√3)-4  =  3k, then find the value of k. 

Solution : 

(91/2)-7 ⋅ (31/2)-4  =  3k

(9)-7/2 ⋅ (3)-4/2  =  3k

(32)-7/2 ⋅ 3-2  =  3k

3⋅ (-7/2) ⋅ 3-2  =  3k

3-7 ⋅ 3-2  =  3k

3-7 - 2  =  3k

3-9  =  3k

k  =  -9

So, the value of k is -9.

Problem 9 :

If √(x√x)   =  xa, then find the value of a.

Solution : 

√(x√x)   =  xa

√(x ⋅ x1/2)   =  xa

√(x1 + 1/2)   =  xa

√(x3/2)   =  xa

(x3/2)1/2   =  xa

x3/4   =  xa

3/4  =  a

So, the value of a is 3/4.

Problem 10 :

If n3   =  x, n4  =  20x and n > 0, then find the value of n. 

Solution : 

n4  =  20x

n⋅ n  =  20x

Substitute x for n3.

  x ⋅ n  =  20x

nx  =  20x

Divide each side by x.

n  =  20

So, the value of n is 20.

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