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Answer:
[tex]\bold{k \: = \: 4}[/tex]
Step-by-step explanation:
[tex]\bold{Given} = > \\ log_{17} \: log_{2}( \: 5 \sqrt{x } - \sqrt{25x - 4} \: ) = 0[/tex]
[tex]the \: value \: of \: x \: satisfying \: the \: above \: equation \: is \: \dfrac{k}{25} [/tex]
[tex]\bold{To \: find }= > value \: of \: k[/tex]
[tex]\{Concept \: used} = > \\ if \: log_{y}(x) \: = p \: then \\ = > x = {y}^{p} [/tex]
[tex]2) {( \sqrt{x}) }^{2} = x[/tex]
[tex]3) \: {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab[/tex]
[tex]\bold{Solution} = > \\ log_{17}( log_{2}(5 \sqrt{x} - \sqrt{25x - 4} ) ) = 0[/tex]
[tex] = > log_{2}(5 \sqrt{x} - \sqrt{25x - 4} ) = {17}^{0} [/tex]
[tex] = > log_{2}(5 \sqrt{x} - \sqrt{25x - 4} ) = 1[/tex]
[tex] = > 5 \sqrt{x} - \sqrt{25x - 4} = {2}^{1} [/tex]
[tex] = > \sqrt{25x - 4} \: = 2 - 5 \sqrt{x} [/tex]
[tex]squaring \: both \: sides \: we \: get[/tex]
[tex] = > { \sqrt{25x - 4} }^{2} = {(2 - 5 \sqrt{x}) }^{2} [/tex]
[tex] = > 25x - 4 \: = {2}^{2} + {(5 \sqrt{x}) }^{2} - 2 \: (2) \: (5 \sqrt{x} )[/tex]
[tex] = > 25x - 4 = 4 + 25x - 20 \sqrt{x} [/tex]
[tex]25x \: is \: cancel \: out \: from \: each \: side[/tex]
[tex] = > 20 \sqrt{x} = 4 + 4[/tex]
[tex] = > \sqrt{x} = \dfrac{8}{20} [/tex]
[tex] = > \sqrt{x} = \dfrac{2}{5} [/tex]
[tex]squaring \: both \: sides[/tex]
[tex] = > {( \sqrt{x} \: )}^{2} = { (\dfrac{2}{5}) }^{2} [/tex]
[tex] = > x \: = \dfrac{4}{25} [/tex]
[tex]now \: it \: is \: given \: that \: solution \: of \: given \: equation \: is \: \\ x \: = \dfrac{k}{25} [/tex]
[tex]so \\ \dfrac{k}{25} = \dfrac{4}{25} [/tex]
[tex] = > \: k \: = 4[/tex]