Find: log_2 x+log_2 2x=4

Find:
${\mathrm{log}}_{2}x+{\mathrm{log}}_{2}2x=4$
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Solution:
${\mathrm{log}}_{2}x+{\mathrm{log}}_{2}2x=4\phantom{\rule{0ex}{0ex}}⇒{\mathrm{log}}_{2}\left(x\cdot 2x\right)=4\phantom{\rule{0ex}{0ex}}⇒{\mathrm{log}}_{2}\left(2{x}^{2}\right)=4\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}={2}^{4}\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}=16\phantom{\rule{0ex}{0ex}}{x}^{2}=\frac{16}{2}=8\phantom{\rule{0ex}{0ex}}x=\sqrt{8}\phantom{\rule{0ex}{0ex}}x=2\sqrt{2}$