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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 byShamim 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 byShamim last updated on 03/Oct/19

hmm. tnks

$$\mathrm{hmm}.\:\mathrm{tnks} \\ $$

Commented byRio Michael last updated on 03/Oct/19

your welcom sir

$${your}\:{welcom}\:{sir} \\ $$

Commented byMJS 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 byMJS 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 byShamim 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 byMJS 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 byShamim last updated on 03/Oct/19

tnks a lot...

$$\mathrm{tnks}\:\mathrm{a}\:\mathrm{lot}... \\ $$

Commented byShamim 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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