Question

$\sf{Solve \: log_{3}( \sqrt{x} + | \sqrt{x} - 1 | ) = log_{9}( 4\sqrt{x} - 3 + 4 | \sqrt{x} - 1 | ) }$
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Answer 1

Answer:

$\sf{ log_{3}( \sqrt{x} + | \sqrt{x} - 1 | ) = log_{9}( 4\sqrt{x} - 3 + 4 | \sqrt{x} - 1 | ) }$.. (i)

From Eq.(i) is defined , ifx≥0

$\sf{then \: log_{3}( \sqrt{x} + | \sqrt{x} - 1 | ) = log_ {{3}^{2}} (4 \sqrt{x} - 3 + 4 | \sqrt{x} - 1| ) }$

$\implies \sf{} \: 2( \sqrt{x} + | \sqrt{x} - 1|) = 4 \sqrt{x} - 3 + 4 | \sqrt{x} - 1 |$

$\implies \sf{}\: 3 - 2 \sqrt{x} = 2 | \sqrt{x} - 1 |$

On squaring both sides, then

$\sf \: \: \: 9 + 4x - 12 \sqrt{x} = 4x + 4 - 4 \sqrt{x}$

$\: \: \: \: \: : \longmapsto \: 8 \sqrt{x} = 5$

$\therefore \: {\underline{\boxed{ \sf{\pink{\: \: \: x = \frac{25}{64}}}}}}$

Answer 2

Answer:

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Step-by-step explanation:

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