Question

If $\sqrt{x^{\frac{1}{x}}.(2x{)}^{\frac{1}{2x}}.(4x{)}^{\frac{1}{4x}}.(8x{)}^{\frac{1}{8x}}\dots \infty }=\frac{32}{3125},$ then $x$ equals

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{5}$
• D. $\frac{1}{6}$

Answer 1

The correct option is C $\frac{1}{5}$

$x^{\frac{1}{2x}}.(2x{)}^{\frac{1}{4x}}.(4x{)}^{\frac{1}{8x}}.(8x{)}^{\frac{1}{16x}}\dots \infty =\frac{32}{3125}$ ... (i)

$LHS=\left(x^{\frac{1}{2x}+\frac{1}{4x}+\frac{1}{8x}+\frac{1}{16x}\dots \infty }\right)\times \left(2^{\frac{1}{4x}+\frac{2}{8x}+\frac{3}{16x}+\dots \infty }\right)=x^{\frac{1}{x}(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots \infty )}\times 2^{\frac{1}{4x}(1+\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+\dots \infty )}$ ... (ii)

Now, let $S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...\infty$
It is an infinite G.P. with first term as $\frac{1}{2}$ and common ratio as $\frac{1}{2}$
So, $S=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$ ... (iii)

And, let$S^{\prime }=1+\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+\dots \infty$ ... (iv)
Multiplying (iv) with $\frac{1}{2}$, we get,
$\frac{1}{2}{S}^{\prime }=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\dots \infty$ ... (v)
On subtracting (v) from (iv), we get,
$\frac{1}{2}{S}^{\prime }=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots \infty$

$\Rightarrow S^{\prime }=2(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots \infty )$

$\Rightarrow S^{\prime }=2\left[\frac{1}{1-\frac{1}{2}}\right]=4$

From (i)

$LHS=x^{\frac{1}{x}\times S}\times 2^{\frac{1}{4x}\times S^{\prime }}=x^{\frac{1}{x}}\times 2^{\frac{1}{x}}=\frac{32}{3125}$

$\Rightarrow (x{)}^{\frac{1}{x}}.(2{)}^{\frac{1}{x}}=\frac{(2{)}^5}{(5{)}^5}={\left(\frac{1}{5}\right)}^5(2{)}^5$
$\Rightarrow x=\frac{1}{5}$