Question

If $4^x-4^{x-1}=24,$ what is the value of ${\left(2x\right)}^x?$

• A. $5\sqrt{5}$
• B. $\sqrt{5}$
• C. $25\sqrt{5}$
• D. 125

Answer 1

The correct option is C $25\sqrt{5}$

We have,

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

$\Rightarrow {\left(2\right)}^{2x}-{\left(2\right)}^{2(x-1)}=24$

$\Rightarrow {\left(2\right)}^{2x}-{\left(2\right)}^{2x-2}=24$

$\Rightarrow {\left(2\right)}^{2x}-{\left(2\right)}^{2x}{\left(2\right)}^{-2}=24$

$\Rightarrow {\left(2\right)}^{2x}-\frac{{\left(2\right)}^{2x}}{4}=24$

$\Rightarrow \frac{4{\left(2\right)}^{2x}-{\left(2\right)}^{2x}}{4}=24$

$\Rightarrow \frac{3}{4}{\left(2\right)}^{2x}=24$

$\Rightarrow {\left(2\right)}^{2x}=8\times 4$

$\Rightarrow {\left(2\right)}^{2x}=32$

$\Rightarrow {\left(2\right)}^{2x}={\left(2\right)}^5$

On comparing that,

$2x=5$

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

Now the value of

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

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

$=25\sqrt{5}$

Hence, this is the answer.