Question

$log_{3} \sqrt{3 + \sqrt{3 + \sqrt{3..... \infty = } } } ?$
$intelligents stundents plzz anwer$
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Answer 1

$Given \: \: Question \: \: Is \: \\ \\ log_{3}( \sqrt{3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + ... \infty } } } } ) \\ \\ let \: \: \: \sqrt{3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + ... \infty } } } } = z \\ \\ remove \: \: \: one \: \: \: \sqrt{3} \\ \\ \sqrt{3 + \sqrt{3 + \sqrt{3 + ... \infty } } } = z \\ \\ squaring \: \: on \: \: both \: \: sides \\ \\ 3 + \sqrt{3 + \sqrt{3 + ... \infty } } = z {}^{2} \\ \\ 3 + z = z {}^{2} \\ \\ z {}^{2} - z - 3 = 0 \\ \\ z = \frac{ - 1 + \sqrt{13} }{1} \: \: \: or \: \: \: z = \frac{ - 1 - \sqrt{13} }{1} \\ \\ therefore \: \: \\ log_{3}( \sqrt{13} - 1)$