Question

if 4^x=4^x-1=24then find the value of (2x)^2x​

Answer 1

Answer:

⇒ $x= \frac{5}{2}$,

So,

$(2x)^{2x} = 3125$

Step-by-step explanation:

Given :

To find the value of :

$(2x)^{2x}$ if,

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

Solution :

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

⇒ $4^{x-1}(4 -1)=24$

⇒ $4^{x-1}(3)=24$

⇒ $4^{x-1}=\frac{24}{3}$

⇒ $4^{x-1}=8$

⇒ $4^{x-1}=2^3$

⇒ $4^{x-1}=[4^{\frac{1}{2}}]^3$

⇒ $4^{x-1}=4^{\frac{3}{2}}$

Since,

Bases are equal, powers must be equal,

⇒ $x - 1 = \frac{3}{2}$

⇒ $x = \frac{3}{2}+1 = \frac{3+2}{2}$

⇒ $x= \frac{5}{2}$

So,

$(2x)^{2x} = (\frac{5}{2}\times2)^{\frac{5}{2}\times 2} = 5^5=3125$

Answer 2

Answer: 3125

Step-by-step explanation:

Given,

$4^x-4^{x-1}=24\\\;\\4^x-4^x.4^{-1}=24\\\;\\\text{Taking}\;\;4^x\;\;\text{Common;}\\\;\\4^x(1-4^{-1})=24\\\;\\4^x(1-\frac{1}{4})=24\\\;\\4^x(\frac{4-1}{4})=24\\\;\\4^x\times\frac{3}{4}=24\\\;\\4^x=\frac{24\times4}{3}\\\;\\4^x=8\times4\\\;\\4^x=32\\\;\\4^x=2^5\\\;\\(2^2)^x=2^5\\\;\\2^{2x}=2^5\\\;\\\text{On comparing base;}\\\;\\2x=5$

Now,

$(2x)^{2x}=5^5\\\;\\(2x)^{2x}=5\times5\times5\times5\times5\\\;\\(2x)^{2x}=3125$

Note:-

$1.)\;\;a^{-1}=\frac{1}{a}\\\;\\2.)\;\;(a^m)^n=a^{mn}$