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y-2-1-log-x-8-then-x-




Question Number 20245 by virus last updated on 24/Aug/17
y=2^(1/(log_x 8))   then x=?
$$\mathrm{y}=\mathrm{2}^{\frac{\mathrm{1}}{{log}_{{x}} \mathrm{8}}} \\ $$$${then}\:{x}=? \\ $$
Answered by Tinkutara last updated on 24/Aug/17
y = 2^(log_8  x)  = 2^((1/3) log_2  x)   y^3  = 2^(log_2  x)  = x  ⇒ x = y^3
$${y}\:=\:\mathrm{2}^{\mathrm{log}_{\mathrm{8}} \:{x}} \:=\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{log}_{\mathrm{2}} \:{x}} \\ $$$${y}^{\mathrm{3}} \:=\:\mathrm{2}^{\mathrm{log}_{\mathrm{2}} \:{x}} \:=\:{x} \\ $$$$\Rightarrow\:{x}\:=\:{y}^{\mathrm{3}} \\ $$
Answered by Rauny last updated on 27/Jan/19
(1/((((ln 8)/(ln x)))))=log_2  y  ((ln x)/(ln 8))=((ln y)/(ln 2))  ln x=((ln 8)/(ln 2))∙ln y=log_2  8∙ln y  =3ln y  x=e^(3ln y)   =y^3   ∴x=y^3
$$\frac{\mathrm{1}}{\left(\frac{{ln}\:\mathrm{8}}{{ln}\:{x}}\right)}={log}_{\mathrm{2}} \:{y} \\ $$$$\frac{{ln}\:{x}}{{ln}\:\mathrm{8}}=\frac{{ln}\:{y}}{{ln}\:\mathrm{2}} \\ $$$${ln}\:{x}=\frac{{ln}\:\mathrm{8}}{{ln}\:\mathrm{2}}\centerdot{ln}\:{y}={log}_{\mathrm{2}} \:\mathrm{8}\centerdot{ln}\:{y} \\ $$$$=\mathrm{3}{ln}\:{y} \\ $$$${x}={e}^{\mathrm{3}{ln}\:{y}} \\ $$$$={y}^{\mathrm{3}} \\ $$$$\therefore{x}={y}^{\mathrm{3}} \\ $$

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