Question

Q1.if 4^x=4^x-1=24then find the value of (2x)^2x​

Answer 1

Step-by-step explanation:

Answer: 3125

Step-by-step explanation:

Given,

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

Now,

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

Note:-

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

Answer 2

Step-by-step explanation:

Answer

Correct option is

D

25

5

4

x

−4

x−1

=24

or 4

x

(1−

4

1

)=24

or 4

x

4

3

=24

or 4

x

=

3

24×4

=32=2

5

or 2

2x

=2

5

or x=

2

5

or 2x=5 or (2x)

x

=(5)

5/2

=5

1

×5

1

×5

1/2

=25

5