Question

can someone answer this question plz ....​

Answer 1

Solution :-

Given :-

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

Now solving it :-

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

$\implies 4^{x-1} \times 4 - 4^{x-1} = 24$

$\implies 4^{x-1} ( 4 - 1) = 24$

$\implies 4^{x-1} (3) = 24$

$\implies 4^{x-1} = 8$

As we can write 4 = 2² and 8 = 2³

$\implies (2^2)^{x-1} = 2^3$

$\implies 2^{2x - 2} = 2^{3}$

As bases are same

$\implies 2x - 2 = 3$

$\implies 2x = 5$

$\implies x = \dfrac{5}{2}$

Now value of $(2x)^{x}$

$= \left(2 \times \dfrac{5}{2} \right)^{\frac{5}{2}}$

$= 5^{\frac{5}{2}}$

$= 25\sqrt{5}$

Answer 2

Answer:

Here 4^x-4^x-1=24

So 4^x-4^x/4^1=24

so taking common 4^x(1-1/4)=24

So 4^x=24*4/3=32

So (2^2)^x=32

So2^2x=32

2x=5

X=5/2

2x^x=2*5/2^5/2=5^5/2=(sq root 5)^5=25*root5