Tolong dong nomor 27 Terimakasih :)

[tex]\displaystyle \log x=6\\x=10^6\\\\\log y = 12\\y=10^{12}\\\\\sqrt{x\sqrt{y\sqrt{x\sqrt{y\sqrt{...}}}}}=\sqrt{10^6\sqrt{10^{12}\sqrt{10^6\sqrt{10^{12}\sqrt{...}}}}}\\\sqrt{x\sqrt{y\sqrt{x\sqrt{y\sqrt{...}}}}}=\sqrt{10^6\cdot10^{6}\sqrt{10^6\cdot10^{6}\sqrt{...}}}}}\\\sqrt{x\sqrt{y\sqrt{x\sqrt{y\sqrt{...}}}}}=\sqrt{10^{12}\sqrt{10^{12}\sqrt{...}}}}}\\\\\text{misal :}\\\sqrt{x\sqrt{y\sqrt{x\sqrt{y\sqrt{...}}}}}=z\\\\z=\sqrt{10^{12}\sqrt{10^{12}\sqrt{...}}}}}\\z^2=10^{12}\sqrt{10^{12}\sqrt{...}}}}[/tex]
[tex]\displaystyle z^2=10^{12}z\\z=10^{12}\\\sqrt{x\sqrt{y\sqrt{x\sqrt{y\sqrt{...}}}}}=10^{12}\\\\\therefore\\\sqrt{\log\sqrt{x\sqrt{y\sqrt{x\sqrt{y\sqrt{...}}}}}}=\sqrt{\log10^{12}}\\\boxed{\boxed{\sqrt{\log\sqrt{x\sqrt{y\sqrt{x\sqrt{y\sqrt{...}}}}}}=6}}[/tex]