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Surds & Indices
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CAT 2025 Lesson : Surds & Indices - Comparing Indices

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2.1 Comparing Indices

To compare indices, we can increase or reduce the powers in the same proportion and compare.

2.1.1 Indices are integers

When the powers are integers, then

(1) Find the HCF of the powers.
(2) Divide each of the powers by the HCF
(3) Arrange the numbers using the new powers obtained in step
22.

Example 7

Which of the following has the largest value?

(1)
2962^{96}     (2) 3723^{72}     (3) 5485^{48}     (4) 232423^{24}

Solution

HCF(96,72,48,24)=24(96, 72, 48, 24) = 24

29624=24=162^\frac{96}{24} = 2^4 = 16

37224=33=273^\frac{72}{24} = 3^3 = 27

54824=52=255^\frac{48}{24} = 5^2 = 25

232424=2323^\frac{24}{24} = 23

2727 is the greatest number.

372\therefore 3^{72} is the largest.

Answer: (2)
3723^{72}

2.1.2 Indices are fractions

When the powers are fractions, then

(1) Ensure the numerator is the same across all powers
(2) Find the LCM of the denominators in the powers.
(3) Multiply each of the powers by the LCM
(4) Order the numbers using the new powers obtained in step 3.

Example 8

Which of the following has the largest value?

(1)
2182^\frac{1}{8}     (2) 31123^\frac{1}{12}     (3) 51165^\frac{1}{16}     (4) 61246^\frac{1}{24}

Solution

LCM(8,12,16,24)=48(8, 12, 16, 24) = 48

218×48=26=642^{\frac{1}{8} \times {48}} = 2^6 = 64

3112×48=34=813^{\frac{1}{12} \times {48}} = 3^4 = 81

5116×48=53=1255^{\frac{1}{16} \times {48}} = 5^{3} = 125

6124×48=62=366^{\frac{1}{24} \times {48}} = 6^{2} = 36

125125 is the greatest number.

5116 \therefore 5^\frac{1}{16} is the largest.

Answer: (3)
51165^{\frac{1}{16}}

2.2 Practice Questions in Indices

In an equation, if the bases are equal, then the powers should also be equal.

Example 9

If 7x+1=49y7^{{x}+{1}} = 49^{y} and 27x=81y×2727^{x} = 81^{y} \times 27, then 2xy=2^{{x}-{y}} =

(1)
22     (2) 44     (3) 88     (4) 1616

Solution

In each of the above equations, the bases are powers of the same number. So, we first make the bases equal.

7x+1=49y7^{{x}+{1}} = 49^{y}
7x+1=72y 7^{{x}+{1}} = 7^{2y}
x+1=2y x + 1 = 2 y
x=2y1 x = 2y - 1 ----- (1)

27x=81y×2727^{x} = 81^{y} \times 27
33x=34y×333^{3x} = 3^{4y} \times 3^3
33x=34y+33^{3x} = 3^{4y + 3}
3x=4y+33x = 4y + 3   ----- (2)

Substituting (1) in (2),

3(2y1)=4y+33(2y - 1) = 4y + 3
2y=62y = 6
y=3,x=5y = 3, x = 5

2xy=253=4 \therefore 2^{x - y} = 2^{5 - 3} = 4

Answer: (2)
44

Example 10

(xaxbc)×(xbxca)2×(xcxab)3=\left( \dfrac{x^{a}}{x^{b - c}} \right) \times \left( \dfrac{x^{b}}{x^{c - a}} \right)^2 \times \left( \dfrac{x^{c}}{x^{a - b}} \right)^3 = ?

(1)
x2a+4cx^{2a + 4c}     (2) x4b+2cx^{4b + 2c}     (3) xa+2b+3cx^{a + 2b + 3c}     (4) xa+b+cx^{a + b + c}

Solution

(xaxbc)×(xbxca)2×(xcxab)3=x(ab+c)×(x(bc+a))2×(x(ca+b))3\left( \dfrac{x^{a}}{x^{b - c}} \right) \times \left( \dfrac{x^{b}}{x^{c - a}} \right)^2 \times \left( \dfrac{x^{c}}{x^{a - b}} \right)^3 = x^{(a - b + c)} \times (x^{(b - c + a)})^2 \times (x^{(c - a + b)})^3

=xab+c+2b2c+2a+3c3a+3b= x^{a - b + c + 2b - 2c + 2a + 3c - 3a + 3b}

=x4b+2c= x^{4b + 2c}

Answer: (2)
x4b+2cx^{4b + 2c}

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