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log-2-x-log-4-y-4-x-y-8-




Question Number 88663 by mary_ last updated on 12/Apr/20
 { ((log_2 x+log_4 y=4)),((x.y=8)) :}
$$\begin{cases}{{log}_{\mathrm{2}} {x}+{log}_{\mathrm{4}} {y}=\mathrm{4}}\\{{x}.{y}=\mathrm{8}}\end{cases} \\ $$
Answered by jagoll last updated on 28/Apr/20
log_2 (x)+(1/2)log_2 ((8/x)) =4  log_2 (x) +log_2  (√(8/x)) = log_2  (16)  x(√(8/x)) = 16 ⇒(√(8x)) = (√(256))  x = ((256)/8) = 32 so y = (8/(32)) = 4^(−1)
$$\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{8}}{\mathrm{x}}\right)\:=\mathrm{4} \\ $$$$\mathrm{log}_{\mathrm{2}} \left({x}\right)\:+\mathrm{log}_{\mathrm{2}} \:\sqrt{\frac{\mathrm{8}}{{x}}}\:=\:\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{16}\right) \\ $$$${x}\sqrt{\frac{\mathrm{8}}{{x}}}\:=\:\mathrm{16}\:\Rightarrow\sqrt{\mathrm{8}{x}}\:=\:\sqrt{\mathrm{256}} \\ $$$${x}\:=\:\frac{\mathrm{256}}{\mathrm{8}}\:=\:\mathrm{32}\:{so}\:{y}\:=\:\frac{\mathrm{8}}{\mathrm{32}}\:=\:\mathrm{4}^{−\mathrm{1}} \\ $$
Commented by john santu last updated on 12/Apr/20
correction sir  it should be   log_2 (x)+log_2  (√(8/x)) = 4  ⇒ x(√(8/x)) = 16 ⇒ (√(8x)) = 16  x = ((16.16)/8) = 32 . then y = (8/(32)) = (1/4)
$${correction}\:{sir} \\ $$$${it}\:{should}\:{be}\: \\ $$$$\mathrm{log}_{\mathrm{2}} \left({x}\right)+\mathrm{log}_{\mathrm{2}} \:\sqrt{\frac{\mathrm{8}}{{x}}}\:=\:\mathrm{4} \\ $$$$\Rightarrow\:{x}\sqrt{\frac{\mathrm{8}}{{x}}}\:=\:\mathrm{16}\:\Rightarrow\:\sqrt{\mathrm{8}{x}}\:=\:\mathrm{16} \\ $$$${x}\:=\:\frac{\mathrm{16}.\mathrm{16}}{\mathrm{8}}\:=\:\mathrm{32}\:.\:{then}\:{y}\:=\:\frac{\mathrm{8}}{\mathrm{32}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by jagoll last updated on 12/Apr/20
oo yes. thank you mr john
$${oo}\:{yes}.\:{thank}\:{you}\:{mr}\:{john} \\ $$

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