SURDS AND INDICES

Surds

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

a^1/n  =  ⁿ√a

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

Laws of Surds

Law 1 : 

ⁿ√a  =  a^1/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)  =  ^mn√a

Law 6 : 

(ⁿ√a)^m  =  ⁿ√a^m

Laws of Indices

Law 1 : 

x^m ⋅ xⁿ  =  x^m+n

Law 2 : 

x^m ÷ xⁿ  =  x^m-n

Law 3 : 

(x^m)ⁿ  =  x^mn

Law 4 : 

(xy)^m  =  x^m ⋅ y^m

Law 5 : 

(x / y)^m  =  x^m / y^m

Law 6 : 

x^-m  =  1 / x^m

Law 7 : 

x⁰  =  1

Law 8 : 

x¹  =  x

Law 9 : 

x^m/n  =  y -----> x  =  y^n/m

Law 10 : 

(x / y)^-m  =  (y / x)^m

Law 11 : 

a^x  =  a^y -----> x  =  y

Law 12 : 

x^a  =  y^a -----> x  =  y

Practice Problems

Problem 1 : 

If x^1/p  =  y^1/q  =  z^1/r and xyz  =  1, then find  the value of
(p + q + r).  

Solution : 

Let  x^1/p  =  y^1/q  =  z^1/r  =  k.

Then,

x^1/p  =  k -----> x  =  k^p

y^1/q  =  k -----> y  =  k^pq

z^1/r  =  k -----> z  =  k^r

Given : xyz  =  1

Then,

xyz  =  1

k^p ⋅ k^q ⋅ k^r  =  1

k^{p + q + r}  =  1 -----(1)

We know that a⁰  =  1.

So,

k⁰ = 1

In (1),  substitute 1  =  k⁰.

(1) -----> k^{p + 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  =  p^1/3 - p^-1/3, find the value of 

x³ + 3x

Solution : 

Given : x = p^1/3 - p^-1/3

Take power 3 on both the sides.

x³  =  (p^1/3 - p^-1/3)³

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

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

x³  =  p - p^-1 - 3p^{1/3 - 1/3}(x)

x³  =  p - 1/p - 3p⁰x

x³  =  p - 1/p - 3(1)x

x³  =  p - 1/p - 3x

x³ + 3x  =  p - 1/p

Problem 4 : 

Simplify : 

Solution : 

Problem 5 : 

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

3x³ - 9x

Solution : 

Given : x = 3^1/3 + 3^-1/3

Take power 3 on both the sides.

x³  =  (3^1/3 + 3^-1/3)³

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

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

x³  =  3 + 3^-1 + 3 ⋅ 3^{1/3 - 1/3} ⋅x

x³  =  3 + 1/3 + 3 ⋅ 3⁰ ⋅ x

x³  =  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 a^x = b, b^y = c and  c^z = a, then find the value of xyz. 

Solution : 

Let

a^x  =  b -----(1)

b^y  =  c -----(2)

c^z  =  a -----(3)

Substitute a  =  c^z in (1).

(1)-----> (c^z)^x  =  b

c^zx  =  b

Substitute c  =  b^y.

(b^y)^zx  =  b

b^xyz  =  b

b^xyz  =  b¹

xyz  =  1

Problem 7 :

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

1/x  +  1/y  +  1/z  

Solution : 

Let 2^x  =  3^y  =  6^-z  =  k.

Then, 

2^x  =  k ----->  2  =  k^1/x

3^y  =  k -----> 3  =  k^1/y

6^-z  =  k -----> 6  =  k^-1/z

And also, 

6  =  k^-1/z 

(2 x 3)  =  k^-1/z

In (1), substitute 2  =  k^1/x, 3  =  k^1/y_.

k^1/x ⋅ k^1/y  =  k^-1/z

k^{1/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  =  3^k, then find the value of k. 

Solution : 

(9^1/2)^-7 ⋅ (3^1/2)^-4  =  3^k

(9)^-7/2 ⋅ (3)^-4/2  =  3^k

(3²)^-7/2 ⋅ 3^-2  =  3^k

3^{2 ⋅ (-7/2)} ⋅ 3^-2  =  3^k

3^-7 ⋅ 3^-2  =  3^k

3^{-7 - 2}  =  3^k

3^-9  =  3^k

k  =  -9

So, the value of k is -9.

Problem 9 :

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

Solution : 

√(x√x)   =  x^a

√(x ⋅ x^1/2)   =  x^a

√(x^{1 + 1/2})   =  x^a

√(x^3/2)   =  x^a

(x^3/2)^1/2   =  x^a

x^3/4   =  x^a

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

n³ ⋅ 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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