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

$$\frac{{(3+2\sqrt{2})}^{2008}}{{(7+5\sqrt{2})}^{1338}}+(3-2\sqrt{2})=\log {}_2\left(\mathrm{x}\right)$$$$\mathrm{x}=?$$

Answered by EDWIN88 last updated on 28/Jul/21

$$\log {}_2\left(x\right)=\frac{{(3+2\sqrt{2})}^{2008}}{{(7+5\sqrt{2})}^{1338}}+(3-2\sqrt{2})$$$$\log {}_2\left(x\right)=\frac{{(1+\sqrt{2})}^{2\times 2008}}{{(3+2\sqrt{2})}^{1338}{(1+\sqrt{2})}^{1338}}+(3-2\sqrt{2})$$$$\log {}_2\left(x\right)=\frac{{(1+\sqrt{2})}^{4016}}{{(1+\sqrt{2})}^{4014}}+(3-2\sqrt{2})$$$$\log {}_2\left(x\right)={(1+\sqrt{2})}^2+3-2\sqrt{2}$$$$\log {}_2\left(x\right)=3+2\sqrt{2}+3-2\sqrt{2}=6$$$$\Rightarrow x=2^6=64$$$$\left[loveJew\right]$$

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