Quiz 3

Given, √15 = 3.88

Now, √(5 / 3) = √ (5 × 3) / (3 × 3) = √15 / 3

= 3.88 / 3 = 1.293

{(6.25 )^1/2 × (0.0144)^1/2 + 1} / {(0.027)^1/3 × (81)^1/4​}

= {(2.5)^2×(1/2) × (0.12)^2×(1/2)+1} / {(0.3)^3×1/3 × (3)^4×(1/4)}

= (2.5 × 0.12 + 1) / (0.3) × (3)

= (0.3 + 1) / 0.9

=1.3/o.9

=1.444

2^2n–1 = 1 / 8^n–3

= 1 / (2³)^n–3

= 1 / 2^3(n–3)

= 1 / 2^(3n–9)

= 2^(9–3n)

Let's take power from the equation:

2n−1 = 9−3n

⇒ 5n = 10

⇒   n = 2.

As 256 = 2⁸;   576 = 24²;   64 = 2⁶;   27 = 3³

(256 / 576)^1/4 × (64 / 27)^–1/3 × (216 / 8)^–1

(2⁸ / 24²)^1/4 × (3³ / 2⁶)^1/3 × (8 / 216)

= (2² / √24) × (3 / 4) × (1 / 27)

= 1 / (2√6 × 9)

= 1 / 18√6

Putting x for (?) , we get

(216 / 1)^–2/3 ÷ (27 / 1)^–4/3 = x

x = (1 / 6)² ÷ (1 / 3)⁴

⇒ x= (1 / 36) ÷ (1 / 81)

⇒ x= (1 / 36) × (81 /1)

⇒ x = (81 / 36) = 9 / 4

√8 + 2 √32 – 3 √128 + 4 √50

= 2 √2 + 8 √2 – 3 × 8 √2 + 4 × 5 √2

= (2 + 8 – 24 + 20) √2

= 6 √2 = 6 × 1.414

= 8.484

\frac{(243)^{\frac{n}{5}}\times 3^{2n+1}}{9^{n}\times 3^{n-1}}

=\frac{(3^5)^{\frac{n}{5}}\times 3^{2n+1}}{3^{2n}\times 3^{n-1}}

=\frac{(3^n)\times 3^{2n+1}}{3^{2n}\times 3^{n-1}}

=\frac{3^{n+2n+1}}{3^{2n+n-1}}

=3^{n+2n+1-(2n+n-1)}

=3^{n+2n+1-2n-n+1}

=3^{2}

=9

=3^{-2}\times 81^{\frac{3}{4}}\div (729)^{\frac{-1}{3}}

=3^{-2}\times (3^4)^{\frac{3}{4}}\div (9^3)^{\frac{-1}{3}}

latex]=\frac{1}{9}\times (3)^{3}\div (9)^{-1}[/latex]

=\frac{1}{9}\times 27\times 9

=27

\left (\left (\sqrt{\frac{16}{25}} \right )^{4x-9}\right )^3=\frac{64}{125}

=>\left (\left (\sqrt{\frac{16}{25}} \right )^{4x-9}\right )^3=\left (\frac{4}{5} \right )^3

=>\left (\sqrt{\frac{16}{25}} \right )^{4x-9}= \frac{4}{5}

=>\left (\frac{4}{5} \right )^{4x-9}= \frac{4}{5}

=>4x - 9 =1

=>4x = 10

=>4x/5=  2