Question

solve this question. if u can solve then i will sure that u will crash Jee Advanced​

Answer 1

Answer:

(c)3√3

Step-by-step explanation:

by soving it,

2^2x=2^3

and i have crashed jee ,i dont need to crash it second time.

Answer 2

Answer:

$3\sqrt{3}$

Step-by-step explanation:

Given :

If $4^{x+1}-4^x=24$

Find , $(2x)^x$

Solution :

$4^{x+1}-4^x=24$

⇒ $4^{x} \times (4) - 4^{x}=24$

⇒ $4^{x}(4-1)=24$

⇒ $3(4^x) = 24$

⇒ $4^x=\frac{24}{3}$

⇒ $4^x = 8$

⇒ $4^x = 2^3$

⇒ $(2^2)^x = 2^3$

⇒ $2^{2x} = 2^3$

As base is same,

The power must be equal,

⇒ $2x = 3$

$x =\frac{3}{2}$ ...(i)

__

So,

⇒ $(2x)^x = (2\times(\frac{3}{2}))^{\frac{3}{2}}$

⇒ $3^{\frac{3}{2}} = (\sqrt{3})^3 = (\sqrt{3})^2 \times \sqrt{3} = 3\sqrt{3}$

EDIT :

It is of class 6 / 7 (or) 8 ,.

NOT TO THE LEVEL OF JEE