Question

The value of ${\log }_{\frac{4}{3}}\left(\frac{56+\sqrt{56+\sqrt{56+\sqrt{56+......\infty }}}}{\sqrt{64\sqrt{64\sqrt{\mathrm{64......}\infty }}}}\right)$ is equal to

• A. 0
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

0

Let $x=\sqrt{56+x}$

$x^2=56+x$

$x^2-x-56=0$

$x^2-8x+7x-56=0$

$x(x-8)+7(x-8)=0$

$(x-8)(x+7)=0$

$x=8;-7$

Let $y=\sqrt{64y}$

$y^2=64y$

$y^2-64y=0$

$y(y-64)=0$

$y=0;y=64$

So the given question can be written as,

$lo{g}_{\frac{4}{3}}\left(\frac{56+x}{y}\right)$

Put $x=-7;y=64$

We get $lo{g}_{\frac{4}{3}}\left(\frac{56-7}{64}\right)$

$={\log }_{\frac{4}{3}}\left(\frac{49}{64}\right)$

$={\log }_{\frac{4}{3}}\left(\frac{7^2}{8^2}\right)$

$=\frac{2}{2}lo{g}_{\frac{4}{3}}\frac{7}{8}$

Put $x=8;y=64$

$=lo{g}_{\frac{4}{3}}\left(\frac{56+8}{64}\right)$

$=lo{g}_{\frac{4}{3}}\left(\frac{64}{64}\right)$

$=lo{g}_{\frac{4}{3}}\left(1\right)$

$=0$

So option A is the correct answer