Question

Solve This !!!!!!!!!!!!!!!

Answer 1

Answer:

$\bold{k \: = \: 4}$

Step-by-step explanation:

$\bold{Given} = > \\ log_{17} \: log_{2}( \: 5 \sqrt{x } - \sqrt{25x - 4} \: ) = 0$

$the \: value \: of \: x \: satisfying \: the \: above \: equation \: is \: \dfrac{k}{25}$

$\bold{To \: find }= > value \: of \: k$

$\{Concept \: used} = > \\ if \: log_{y}(x) \: = p \: then \\ = > x = {y}^{p}$

$2) {( \sqrt{x}) }^{2} = x$

$3) \: {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab$

$\bold{Solution} = > \\ log_{17}( log_{2}(5 \sqrt{x} - \sqrt{25x - 4} ) ) = 0$

$= > log_{2}(5 \sqrt{x} - \sqrt{25x - 4} ) = {17}^{0}$

$= > log_{2}(5 \sqrt{x} - \sqrt{25x - 4} ) = 1$

$= > 5 \sqrt{x} - \sqrt{25x - 4} = {2}^{1}$

$= > \sqrt{25x - 4} \: = 2 - 5 \sqrt{x}$

$squaring \: both \: sides \: we \: get$

$= > { \sqrt{25x - 4} }^{2} = {(2 - 5 \sqrt{x}) }^{2}$

$= > 25x - 4 \: = {2}^{2} + {(5 \sqrt{x}) }^{2} - 2 \: (2) \: (5 \sqrt{x} )$

$= > 25x - 4 = 4 + 25x - 20 \sqrt{x}$

$25x \: is \: cancel \: out \: from \: each \: side$

$= > 20 \sqrt{x} = 4 + 4$

$= > \sqrt{x} = \dfrac{8}{20}$

$= > \sqrt{x} = \dfrac{2}{5}$

$squaring \: both \: sides$

$= > {( \sqrt{x} \: )}^{2} = { (\dfrac{2}{5}) }^{2}$

$= > x \: = \dfrac{4}{25}$

$now \: it \: is \: given \: that \: solution \: of \: given \: equation \: is \: \\ x \: = \dfrac{k}{25}$

$so \\ \dfrac{k}{25} = \dfrac{4}{25}$

$= > \: k \: = 4$