Question Number 148494 by liberty last updated on 28/Jul/21
$$\:\:\:\frac{\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2008}} }{\left(\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{1338}} }\:+\:\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:=\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{x}\right) \\ $$$$\:\mathrm{x}=?\: \\ $$
Answered by EDWIN88 last updated on 28/Jul/21
![log _2 (x)=(((3+2(√2))^(2008) )/((7+5(√2))^(1338) )) +(3−2(√2)) log _2 (x)=(((1+(√2))^(2×2008) )/((3+2(√2))^(1338) (1+(√2))^(1338) ))+(3−2(√2)) log _2 (x)=(((1+(√2))^(4016) )/((1+(√2))^(4014) )) +(3−2(√2)) log _2 (x)=(1+(√2))^2 +3−2(√2) log _2 (x)=3+2(√2)+3−2(√2) =6 ⇒x = 2^6 = 64 [ love Jew ]](https://www.tinkutara.com/question/Q148495.png)
$$\:\mathrm{log}\:_{\mathrm{2}} \left({x}\right)=\frac{\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2008}} }{\left(\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{1338}} }\:+\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left({x}\right)=\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}×\mathrm{2008}} }{\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{1338}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{1338}} }+\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left({x}\right)=\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{4016}} }{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{4014}} }\:+\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left({x}\right)=\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\: \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left({x}\right)=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\:=\mathrm{6} \\ $$$$\Rightarrow{x}\:=\:\mathrm{2}^{\mathrm{6}} \:=\:\mathrm{64}\: \\ $$$$\:\:\:\:\left[\:{love}\:{Jew}\:\right]\: \\ $$