Question Number 70270 by Shamim last updated on 02/Oct/19
$$\mathrm{If}\:\mathrm{log}_{\mathrm{x}} \mathrm{y}\:=\:\mathrm{6}\:\&\:\mathrm{log}_{\mathrm{14x}} \mathrm{8y}\:=\:\mathrm{3}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\&\:\mathrm{y}. \\ $$
Answered by Rio Michael last updated on 02/Oct/19
$${log}_{{x}} {y}\:=\:\mathrm{6}\:………..\left(\mathrm{1}\right) \\ $$$${log}_{\mathrm{14}{x}} \mathrm{8}{y}\:=\:\mathrm{3}\:……….\left(\mathrm{2}\right) \\ $$$${from}\:{eqn}\:\left(\mathrm{1}\right)\:,\:{y}\:=\:{x}^{\mathrm{6}} …….\left(\mathrm{3}\right) \\ $$$$\:{from}\:{eqn}\left(\mathrm{2}\right),\:\mathrm{8}{y}\:=\:\left(\mathrm{14}{x}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{8}{y}\:=\:\mathrm{2744}\:{x}^{\mathrm{3}} \:…….\left(\mathrm{4}\right) \\ $$$${substitute}\:{eqn}\left(\mathrm{3}\right)\:{in}\:{eqn}\left(\mathrm{4}\right) \\ $$$$\Rightarrow\:\mathrm{8}{x}^{\mathrm{6}} \:=\:\mathrm{2744}\:{x}^{\mathrm{3}} \\ $$$$\:\:\mathrm{8}{x}^{\mathrm{3}} \:=\:\mathrm{2744} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{7} \\ $$$${x}\:=\:\mathrm{7}\:\:\Rightarrow\:{y}\:=\:\mathrm{7}^{\mathrm{6}\:} \\ $$$${x}\:=\:\mathrm{7}\:\:{and}\:{y}\:=\:\mathrm{7}^{\mathrm{6}} \\ $$
Commented by Shamim last updated on 03/Oct/19
$$\mathrm{Here},\:\mathrm{x}\:\mathrm{is}\:\mathrm{base}\:\mathrm{of}\:\mathrm{logaritham}\:\&\:\mathrm{y}\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\mathrm{normal}\:\mathrm{number}.\:\mathrm{bt}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{that}\:\mathrm{y}\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\mathrm{power}\:\mathrm{of}\:\mathrm{x}. \\ $$
Commented by Shamim last updated on 03/Oct/19
$$\mathrm{hmm}.\:\mathrm{tnks} \\ $$
Commented by Rio Michael last updated on 03/Oct/19
$${your}\:{welcom}\:{sir} \\ $$
Commented by MJS last updated on 03/Oct/19
$$\mathrm{log}_{{x}} \:{y}\:=\frac{\mathrm{log}\:{y}}{\mathrm{log}\:{x}}=\mathrm{6} \\ $$$$\mathrm{log}_{\mathrm{14}{x}} \:\mathrm{8}{y}\:=\frac{\mathrm{log}\:{y}\:+\mathrm{3log}\:\mathrm{2}}{\mathrm{log}\:{x}\:+\mathrm{log}\:\mathrm{7}\:+\mathrm{log}\:\mathrm{2}}=\mathrm{3} \\ $$$$ \\ $$$$\mathrm{log}\:{y}\:=\mathrm{6log}\:{x} \\ $$$$\mathrm{log}\:{y}\:=\mathrm{3}\left(\mathrm{log}\:{x}\:+\mathrm{log}\:\mathrm{7}\right) \\ $$$$ \\ $$$$\mathrm{6log}\:{x}\:=\mathrm{3}\left(\mathrm{log}\:{x}\:+\mathrm{log}\:\mathrm{7}\right) \\ $$$$\mathrm{log}\:{x}\:=\mathrm{log}\:\mathrm{7} \\ $$$${x}=\mathrm{7} \\ $$$$\mathrm{log}\:{y}\:=\mathrm{6log}\:\mathrm{7} \\ $$$${y}=\mathrm{7}^{\mathrm{6}} \\ $$
Commented by MJS last updated on 03/Oct/19
$$\left(\mathrm{1}\right)\:\:{z}={x}^{{y}} \:\Leftrightarrow\:{y}=\mathrm{log}_{{x}} \:{z} \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{log}\:{z}\:=\mathrm{log}\:{x}^{{y}} \\ $$$$\mathrm{log}\:{z}\:={y}\mathrm{log}\:{x}\:\Rightarrow\:{y}=\frac{\mathrm{log}\:{z}}{\mathrm{log}\:{x}} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{log}_{{x}} \:{z}\:=\frac{\mathrm{log}\:{z}}{\mathrm{log}\:{x}} \\ $$
Commented by Shamim last updated on 03/Oct/19
$$\mathrm{tnks}\:\mathrm{a}\:\mathrm{lot}..\:\mathrm{bt}\:\mathrm{one}\:\mathrm{prblm} \\ $$
Commented by MJS last updated on 03/Oct/19
$$\mathrm{just}\:\mathrm{change}\:\mathrm{the}\:\mathrm{names}\:\mathrm{of}\:\mathrm{the}\:\mathrm{variables} \\ $$$${z}=\mathrm{log}_{{x}} \:{y}\:\Leftrightarrow\:{x}^{{z}} ={y} \\ $$$$\mathrm{log}\:{x}^{{z}} \:=\mathrm{log}\:{y}\:\Rightarrow\:{z}\mathrm{log}\:{x}\:=\mathrm{log}\:{y}\:\Rightarrow\:{z}=\frac{\mathrm{log}\:{y}}{\mathrm{log}\:{x}} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{log}_{{x}} \:{y}\:=\frac{\mathrm{log}\:{y}}{\mathrm{log}\:{x}} \\ $$
Commented by Shamim last updated on 03/Oct/19
$$\mathrm{tnks}\:\mathrm{a}\:\mathrm{lot}… \\ $$
Commented by Shamim last updated on 03/Oct/19
$$\mathrm{plz}\:\mathrm{give}\:\mathrm{me}\:\mathrm{the}\:\mathrm{law}−−− \\ $$$$ \\ $$$$\mathrm{log}_{\mathrm{x}} \mathrm{y}=\frac{\mathrm{log}\:\mathrm{y}}{\mathrm{log}\:\mathrm{x}}….\:\mathrm{plz}\:\mathrm{describe}\:\mathrm{details}. \\ $$