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If-log-x-y-6-amp-log-14x-8y-3-then-find-the-value-of-x-amp-y-




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

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