[tex] \sf{Solve \: log_{3}( \sqrt{x} + | \sqrt{x} - 1 | ) = log_{9}( 4\sqrt{x} - 3 + 4 | \sqrt{x} - 1 | ) }[/tex]
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[tex] \sf{Solve \: log_{3}( \sqrt{x} + | \sqrt{x} - 1 | ) = log_{9}( 4\sqrt{x} - 3 + 4 | \sqrt{x} - 1 | ) }[/tex]
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[tex] \sf{ log_{3}( \sqrt{x} + | \sqrt{x} - 1 | ) = log_{9}( 4\sqrt{x} - 3 + 4 | \sqrt{x} - 1 | ) }[/tex].. (i)
From Eq.(i) is defined , ifx≥0
[tex] \sf{then \: log_{3}( \sqrt{x} + | \sqrt{x} - 1 | ) = log_ {{3}^{2}} (4 \sqrt{x} - 3 + 4 | \sqrt{x} - 1| ) }[/tex]
[tex] \implies \sf{} \: 2( \sqrt{x} + | \sqrt{x} - 1|) = 4 \sqrt{x} - 3 + 4 | \sqrt{x} - 1 | [/tex]
[tex] \implies \sf{}\: 3 - 2 \sqrt{x} = 2 | \sqrt{x} - 1 | [/tex]
On squaring both sides, then
[tex] \sf \: \: \: 9 + 4x - 12 \sqrt{x} = 4x + 4 - 4 \sqrt{x} [/tex]
[tex] \: \: \: \: \: : \longmapsto \: 8 \sqrt{x} = 5[/tex]
[tex] \therefore \: {\underline{\boxed{ \sf{\pink{\: \: \: x = \frac{25}{64}}}}}}[/tex]
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