If x = 3 - √ {5,} then {{√ x } over {√ 2 + √ {(3x

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Basic of Logarithms

hard

If $x = 3 - \sqrt {5,} $ then ${{\sqrt x } \over {\sqrt 2 + \sqrt {(3x - 2)} }} = $

$5$

$\sqrt 5 $

$1/5$

$1/\sqrt 5 $

Solution

(d) $x = 3 – \sqrt 5 $

$\sqrt x = \sqrt {3 – \sqrt 5 } = {1 \over {\sqrt 2 }}\,.\sqrt {6 – 2\sqrt 5 } = {1 \over {\sqrt 2 }}(\sqrt 5 – 1)$ $3x – 2 = 9 – 3\sqrt 5 – 2 = 7 – 3\sqrt 5 = {{14 – 6\sqrt 5 } \over 2}$

= ${{{{(3 – \sqrt 5 )}^2}} \over 2}$;

$ \Rightarrow $ $\sqrt 2 + \sqrt {3x – 2} = \sqrt 5 \,.\,\sqrt x $;

$\therefore {{\sqrt x } \over {\sqrt 2 + \sqrt {3x – 2} }} = {1 \over {\sqrt 5 }}$.

Standard 11

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Solution

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