Question Number 88663 by mary_ last updated on 12/Apr/20
$$\begin{cases}{{log}_{\mathrm{2}} {x}+{log}_{\mathrm{4}} {y}=\mathrm{4}}\\{{x}.{y}=\mathrm{8}}\end{cases} \\ $$
Answered by jagoll last updated on 28/Apr/20
$$\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{8}}{\mathrm{x}}\right)\:=\mathrm{4} \\ $$$$\mathrm{log}_{\mathrm{2}} \left({x}\right)\:+\mathrm{log}_{\mathrm{2}} \:\sqrt{\frac{\mathrm{8}}{{x}}}\:=\:\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{16}\right) \\ $$$${x}\sqrt{\frac{\mathrm{8}}{{x}}}\:=\:\mathrm{16}\:\Rightarrow\sqrt{\mathrm{8}{x}}\:=\:\sqrt{\mathrm{256}} \\ $$$${x}\:=\:\frac{\mathrm{256}}{\mathrm{8}}\:=\:\mathrm{32}\:{so}\:{y}\:=\:\frac{\mathrm{8}}{\mathrm{32}}\:=\:\mathrm{4}^{−\mathrm{1}} \\ $$
Commented by john santu last updated on 12/Apr/20
$${correction}\:{sir} \\ $$$${it}\:{should}\:{be}\: \\ $$$$\mathrm{log}_{\mathrm{2}} \left({x}\right)+\mathrm{log}_{\mathrm{2}} \:\sqrt{\frac{\mathrm{8}}{{x}}}\:=\:\mathrm{4} \\ $$$$\Rightarrow\:{x}\sqrt{\frac{\mathrm{8}}{{x}}}\:=\:\mathrm{16}\:\Rightarrow\:\sqrt{\mathrm{8}{x}}\:=\:\mathrm{16} \\ $$$${x}\:=\:\frac{\mathrm{16}.\mathrm{16}}{\mathrm{8}}\:=\:\mathrm{32}\:.\:{then}\:{y}\:=\:\frac{\mathrm{8}}{\mathrm{32}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by jagoll last updated on 12/Apr/20
$${oo}\:{yes}.\:{thank}\:{you}\:{mr}\:{john} \\ $$