3-2-2-2008-7-5-2-1338-3-2-2-log-2-x-x-

Question Number 148494 by liberty last updated on 28/Jul/21

\frac{{(3 + 2 \sqrt{2})}^{2008}}{{(7 + 5 \sqrt{2})}^{1338}} + (3 - 2 \sqrt{2}) = \log_{2} (x)

x = ?

Answered by EDWIN88 last updated on 28/Jul/21

\log_{2} (x) = \frac{{(3 + 2 \sqrt{2})}^{2008}}{{(7 + 5 \sqrt{2})}^{1338}} + (3 - 2 \sqrt{2})

\log_{2} (x) = \frac{{(1 + \sqrt{2})}^{2 \times 2008}}{{(3 + 2 \sqrt{2})}^{1338} {(1 + \sqrt{2})}^{1338}} + (3 - 2 \sqrt{2})

\log_{2} (x) = \frac{{(1 + \sqrt{2})}^{4016}}{{(1 + \sqrt{2})}^{4014}} + (3 - 2 \sqrt{2})

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

\log_{2} (x) = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} = 6

\Rightarrow x = 2^{6} = 64

[l o v e J e w]

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