Number systems — Maths STD 9 — Question

We have to find the value of $(2\text{x})^\text{x}$ if $4\text{x}-4^{\text{x}-1}=24$
So,
Taking $4x$ as common factor we get
$4\text{x}(1-4^{-1})=24$
$4\text{x}\Big(1-\frac{1}{4}\Big)=24$
$4\text{x}\Big(\frac{1\times4}{1\times4}-\frac{1}{4}\Big)=24$
$4^4\Big(\frac{4-1}{4}\Big)=24$
$4^\text{x}\times\frac{3}{4}=24$
$4^\text{x}=24\times\frac{4}{3}$
$4^\text{x}=32$
$2^{2\text{x}}=2^{5}$
By equating powers of exponents we get
$2\text{x}=5$
$\text{x}=\frac{5}{2}$
By substituting $\text{x}=\frac{5}{2}$ in $(2\text{x})^\text{x}$ we get
$(2\text{x})^\text{x}=\Big(2\times\frac{5}{2}\Big)^{\frac{5}{2}}$
$=\Big(2\times\frac{5}{2}\Big)^{\frac{5}{2}}$
$=5^{\frac{5}{2}}$
$=5^{5\times\frac{1}{2}}$
$(2\text{x})^\text{x}=\sqrt[2]{5^5}$
$=\sqrt[2]{5\times5\times5\times5}$
$=5\times5\times^\sqrt[2]{5}$
$=25\sqrt{5}$
Hence the correct choice is $c.$