CRE 1 - Indices

Given, $3^{\frac{1}{2}}\times 3^{\frac{3}{2}}\times 3^{\frac{5}{2}}\times 3^{\frac{7}{2}}\times 3^{\frac{9}{2}}\times 3^{\frac{11}{\mathrm{2 }}}$= 27^x

⇒ $3^{\frac{1}{2}+\frac{3}{2}+\frac{5}{2}+\frac{7}{2}+\frac{9}{2}+\frac{11}{\mathrm{2 }}}$= 27^x

⇒ $3^{\frac{1+3+5+7+9+11}{2}}$ = (3³)^x

⇒ 3^36/2 = 3^3x

⇒ 3¹⁸ = 3^3x

⇒ 3x = 18

⇒ x = 6

Hence, option (a).

Given, $3^{x^x}$ = 81

⇒ $3^{x^x}$ = 3⁴

⇒ x^x = 4 = 2²

∴ x = 2.

Hence, 2.

Given, ${\left[{\left(\frac{2}{3}\right)}^2\right]}^3\times {\left(\frac{1}{3}\right)}^{-4}\times 3^{-1}\times \frac{1}{6}$

= ${\left(\frac{2}{3}\right)}^{2\times 3}\times 3^4\times \frac{1}{3}\times \frac{1}{6}$

= $\frac{2^6}{3^6}\times 3^4\times \frac{1}{3}\times \frac{1}{2\times 3}$

= $\frac{2^5}{3^4}$ = $\frac{32}{81}$

Hence, option (b).

Given, x^z = y² 

⇒ 10^{(0.5 × z)} = 10^{(2 × 0.7)}

⇒ 10^0.5z = 10^1.4

⇒ 0.5z = 1.4

⇒ z = 1.4/0.5 = 2.8

Hence, option (c).

2^5x-1 = $2^{2^{4^{1^{5^{2^{9^2}}}}}}$ = $2^{2^{4^1}}$ = $2^{2^4}$ = 2¹⁶

⇒ 5x - 1 = 16

⇒ x = 17/5

Hence, option (d).

Given, 15^3x+3 = 225^x-1

⇒ 15^3x+3 = (15²)^x-1 = 15^2x-2

⇒ 3x + 3 = 2x - 2

⇒ x = -5

Hence, option (d).

Given, 2^a3^b = 864

864 can be factorized as 2⁵ × 3³.

∴ 2^a × 3^b = 2⁵ × 3³

⇒ a = 5 and b = 3.

∴ a + b = 5 + 3 = 8

Hence, 8.

Given, ${\left(5\sqrt{7}\right)}^{5\mathrm{x}-4}$ = ${\left(35\right)}^3{\left(25\right)}^{3/2}$

⇒ $5^{5x-4}\times {\left(7^{\frac{1}{2}}\right)}^{5x-4}$ = $5^3\times 7^3\times {\left(5^2\right)}^{3/2}$

⇒ $5^{5x-4}\times 7^{\frac{5x-4}{2}}$ = $5^6\times 7^3$

Since 5 and 7 are prime numbers, power of 5 should be same in LHS and RHS. (Same can be solved w.r.t. power of 7)

⇒ 5x – 4 = 6

⇒ x = 2

Hence, option (a).

Given: ${\left(\frac{{\mathrm{a}}^{\mathrm{x}+\mathrm{y}}}{a^y}\right)}^{\frac{x}{yz}}\times {\left(\frac{{\mathrm{a}}^{\mathrm{y}+\mathrm{z}}}{a^z}\right)}^{\frac{y}{zx}}\times {\left(\frac{{\mathrm{a}}^{\mathrm{z}+\mathrm{x}}}{a^x}\right)}^{\frac{z}{yx}}$

Considering ${\left(\frac{{\mathrm{a}}^{\mathrm{x}+\mathrm{y}}}{a^y}\right)}^{\frac{x}{yz}}$

⇒ ${\left(\frac{{\mathrm{a}}^{\mathrm{x}+\mathrm{y}}}{a^y}\right)}^{\frac{x}{yz}}={\left(a^{x+y-y}\right)}^{\frac{x}{yz}}$ = ${\left(a^x\right)}^{\frac{x}{yz}}=a^{\frac{x^2}{yz}}$

Similarly:

⇒ ${\left(\frac{{\mathrm{a}}^{\mathrm{y}+\mathrm{z}}}{a^z}\right)}^{\frac{y}{zx}}=a^{\frac{y^2}{zx}}$

⇒ ${\left(\frac{{\mathrm{a}}^{\mathrm{z}+\mathrm{x}}}{a^x}\right)}^{\frac{z}{yx}}=a^{\frac{z^2}{xy}}$

Hence,

${\left(\frac{{\mathrm{a}}^{\mathrm{x}+\mathrm{y}}}{a^y}\right)}^{\frac{x}{yz}}\times {\left(\frac{{\mathrm{a}}^{\mathrm{y}+\mathrm{z}}}{a^z}\right)}^{\frac{y}{zx}}\times {\left(\frac{{\mathrm{a}}^{\mathrm{z}+\mathrm{x}}}{a^x}\right)}^{\frac{z}{yx}}$ 

= $\left(a^{\frac{x^2}{yz}}\right)\times \left(a^{\frac{y^2}{zx}}\right)\times \left(a^{\frac{z^2}{xz}}\right)$

= $a^{\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xz}}$ = $a^{\frac{x^3+y^3+z^3}{xyz}}$ = $a^{\frac{3xyz}{xyz}}$ = $a^3$

Note: a³ + b³ + c³ = 3abc when a + b + c = 0.

Hence, option (b).

We have, 2³⁰⁰ = (2³)¹⁰⁰ = 8¹⁰⁰

Similarly, 3²⁰⁰ = (3²)¹⁰⁰ = 9¹⁰⁰

Since the power for both numbers is same i.e., 100, number with bigger base will be bigger.

Hence, option (b).

Given, 3^1/2, 4^1/5, 5^1/4

LCM (2, 5, 4) = 20

Raising all the numbers to power 20, we get:

⇒ ${\left(3^{\frac{1}{2}}\right)}^{20}, {\left(4^{\frac{1}{5}}\right)}^{20}, {\left(5^{\frac{1}{4}}\right)}^{20}$

⇒ 3¹⁰, 4⁴, 5⁵

⇒ 9⁵, 256, 3125

⇒ 81 × 729, 256, 3125

Since, 81 × 729 > 3125 > 256

⇒ 3^1/2 > 5^1/4 > 4^1/5

Hence, option (d).

Given, $\sqrt{\mathrm{x}\sqrt{\mathrm{x}\sqrt{\mathrm{x}\dots \dots \mathrm{\infty }}}}={11}^{\mathrm{x}}$

Let’s assume √$\sqrt{\mathrm{x}\sqrt{\mathrm{x}\sqrt{\mathrm{x}\dots \dots \mathrm{\infty }}}}$ = y

⇒ √xy = y

⇒ xy = y²

⇒ x = y

Hence, 

$\sqrt{\mathrm{x}\sqrt{\mathrm{x}\sqrt{\mathrm{x}\dots \dots \mathrm{\infty }}}}$ = x = 11^x

⇒ x = 11^1/x

⇒ $\sqrt[x]{x}$ = 11

Hence, option (a).

Let y = $\sqrt{2+\sqrt{2+\sqrt{2+\mathrm{ }\dots \dots \mathrm{\infty }}}}$ 

⇒ y = $\sqrt{2+\mathrm{y}}$

⇒ y² = 2 + y

⇒ y² – y – 2 = 0

⇒ y² – 2y + y – 2 = 0

⇒ (y - 2)(y + 1) = 0

Hence, y = 2 or -1

Since $\sqrt{2+\sqrt{2+\sqrt{2+\mathrm{ }\dots \dots \mathrm{\infty }}}}$ cannot be negative.

Hence, $\sqrt{2+\sqrt{2+\sqrt{2+\mathrm{ }\dots \dots \mathrm{\infty }}}}$ = 2

Hence, option (a).

Let 5^x = (0.125)^y = 10^3z = k

Hence, 5 = k^1/x

0.125 = k^1/y and 

10 = k^1/3z

We know that 0.125 = (0.5)³ = (5/10)³

Hence,  $k^{\frac{1}{y}} = {\left(\frac{k^{\frac{1}{x}}}{k^{\frac{1}{3z}}}\right)}^3=k^{\frac{3}{x}-\frac{1}{z}}$

⇒ $\frac{1}{y}=\frac{3}{x}-\frac{1}{z}$

⇒ $\frac{3}{x}=\frac{1}{y}+\frac{1}{z}$

Hence, option (b).

Let 3^x = 2^y = 6^z = k

Now, 3x = k ⇒ 3 = k^1/x   …(1)

Similarly, 

2y = k ⇒ 2 = k^1/y   …(2)

6y = k ⇒ 6 = k^1/z   …(3)

We know, 3 × 2 = 6

∴ (1) × (2) = (3)

⇒ k^1/x × k^1/y = k^1/z

⇒ $k^{\frac{1}{x}+\frac{1}{y}} = k^{\frac{1}{z}}$

⇒ $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$

⇒ $z=\frac{xy}{x+y}$

Hence, option (b).

Given, $\left(\sqrt[-\mathrm{pq}]{{\mathrm{a}}^{\mathrm{r}}}\right)\times \left(\sqrt[-\mathrm{qr}]{{\mathrm{a}}^{\mathrm{p}}}\right)\times \left(\sqrt[-\mathrm{pr}]{{\mathrm{a}}^{\mathrm{q}}}\right)$

Let’s consider $\sqrt[-\mathrm{pq}]{{\mathrm{a}}^{\mathrm{r}}}$.

⇒ $\sqrt[-\mathrm{pq}]{{\mathrm{a}}^{\mathrm{r}}}$ = ${\left({\mathrm{a}}^{\mathrm{r}}\right)}^{-\frac{1}{pq}}$ = ${\mathrm{a}}^{-\frac{\mathrm{r}}{\mathrm{pq}}}$   …(1)

Similarly, 

$\sqrt[-\mathrm{qr}]{{\mathrm{a}}^p}$ = ${\mathrm{a}}^{-\frac{\mathrm{p}}{\mathrm{qr}}}$   …(2)

$\sqrt[-\mathrm{rp}]{{\mathrm{a}}^q}$ = ${\mathrm{a}}^{-\frac{q}{\mathrm{rp}}}$   …(3)

Multiplying all three equation,

$\left(\sqrt[-\mathrm{pq}]{{\mathrm{a}}^{\mathrm{r}}}\right)\times \left(\sqrt[-\mathrm{qr}]{{\mathrm{a}}^{\mathrm{p}}}\right)\times \left(\sqrt[-\mathrm{pr}]{{\mathrm{a}}^{\mathrm{q}}}\right)$ = ${\mathrm{a}}^{-\frac{\mathrm{r}}{\mathrm{pq}}}\times {\mathrm{a}}^{-\frac{\mathrm{p}}{\mathrm{qr}}}\times {\mathrm{a}}^{-\frac{q}{\mathrm{rp}}}$

= ${\mathrm{a}}^{-\frac{\mathrm{r}}{\mathrm{pq}}}\times {\mathrm{a}}^{-\frac{\mathrm{p}}{\mathrm{qr}}}\times {\mathrm{a}}^{-\frac{q}{\mathrm{rp}}}$

= ${\mathrm{a}}^{-\left(\frac{{\mathrm{r}}^2+p^2+q^2}{\mathrm{pqr}}\right)}$

= ${\mathrm{a}}^{-\left(\frac{1}{\mathrm{pqr}}\right)}$ [p² + q² + r² = 1]

Hence, option (d).