CAT 2025 Lesson : Surds & Indices - Comparing Indices

Which of the following has the largest value?

(1) $2^{96}$     (2) $3^{72}$     (3) $5^{48}$     (4) $23^{24}$

Solution

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

$2^\frac{96}{24} = 2^4 = 16$

$3^\frac{72}{24} = 3^3 = 27$

$5^\frac{48}{24} = 5^2 = 25$

$23^\frac{24}{24} = 23$

$27$ is the greatest number.

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

Answer: (2) $3^{72}$

Which of the following has the largest value?

(1) $2^\frac{1}{8}$     (2) $3^\frac{1}{12}$     (3) $5^\frac{1}{16}$     (4) $6^\frac{1}{24}$

Solution

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

$2^{\frac{1}{8} \times {48}} = 2^6 = 64$

$3^{\frac{1}{12} \times {48}} = 3^4 = 81$

$5^{\frac{1}{16} \times {48}} = 5^{3} = 125$

$6^{\frac{1}{24} \times {48}} = 6^{2} = 36$

$125$ is the greatest number.

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

Answer: (3) $5^{\frac{1}{16}}$

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

(1) $2$     (2) $4$     (3) $8$     (4) $16$

Solution

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

$7^{{x}+{1}} = 49^{y}$
⇒ $7^{{x}+{1}} = 7^{2y}$
⇒ $x + 1 = 2 y$
⇒ $x = 2y - 1$ ----- (1)

$27^{x} = 81^{y} \times 27$
⇒ $3^{3x} = 3^{4y} \times 3^3$
⇒ $3^{3x} = 3^{4y + 3}$
⇒ $3x = 4y + 3$   ----- (2)

Substituting (1) in (2),

$3(2y - 1) = 4y + 3$
⇒ $2y = 6$
⇒ $y = 3, x = 5$

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

Answer: (2) $4$

$\left( \dfrac{x^{a}}{x^{b - c}} \right) \times \left( \dfrac{x^{b}}{x^{c - a}} \right)^2 \times \left( \dfrac{x^{c}}{x^{a - b}} \right)^3 =$ ?

(1) $x^{2a + 4c}$     (2) $x^{4b + 2c}$     (3) $x^{a + 2b + 3c}$     (4) $x^{a + b + c}$

Solution

$\left( \dfrac{x^{a}}{x^{b - c}} \right) \times \left( \dfrac{x^{b}}{x^{c - a}} \right)^2 \times \left( \dfrac{x^{c}}{x^{a - b}} \right)^3 = x^{(a - b + c)} \times (x^{(b - c + a)})^2 \times (x^{(c - a + b)})^3$

$= x^{a - b + c + 2b - 2c + 2a + 3c - 3a + 3b}$

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

Answer: (2) $x^{4b + 2c}$