x-log-3-2-x-1-

Question Number 77631 by Maclaurin Stickker last updated on 08/Jan/20

x^{{l o g}_{3} (2)} = \sqrt{x} + 1

Answered by mr W last updated on 08/Jan/20

\log_{3} 2 = \log_{x} (\sqrt{x} + 1)

\frac{\ln 2}{\ln 3} = \frac{\ln (\sqrt{x} + 1)}{\ln x}

\frac{\ln 2^{n}}{\ln 3^{n}} = \frac{\ln (\sqrt{x} + 1)}{\ln x}

\Rightarrow x = 3^{n}

\Rightarrow \sqrt{x} + 1 = 2^{n}

w e s e e t h e o n l y s o l u t i o n i s n = 2 :

x = 3^{2} = 9

\sqrt{x} + 1 = \sqrt{9} + 1 = 4 = 2^{2} \Rightarrow o k

t h e r e f o r e x = 9 .

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