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Question Number 70270 by Shamim last updated on 02/Oct/19

$$\mathrm{If}{\log }_{\mathrm{x}}\mathrm{y}=6\mathrm{\&}{\log }_{14\mathrm{x}}8\mathrm{y}=3\mathrm{then}\mathrm{find}\mathrm{the}$$$$\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{\&}\mathrm{y}.$$

Answered by Rio Michael last updated on 02/Oct/19

$${log}_{x}y=6...........\left(1\right)$$$${log}_{14x}8y=3..........\left(2\right)$$$$fromeqn\left(1\right),y=x^6.......\left(3\right)$$$$fromeqn\left(2\right),8y={\left(14x\right)}^3$$$$8y=2744x^3.......\left(4\right)$$$$substituteeqn\left(3\right)ineqn\left(4\right)$$$$\Rightarrow 8x^6=2744x^3$$$$8x^3=2744$$$$\Rightarrow x=7$$$$x=7\Rightarrow y=7^6$$$$x=7andy=7^6$$

Commented by Shamim last updated on 03/Oct/19

$$\mathrm{Here},\mathrm{x}\mathrm{is}\mathrm{base}\mathrm{of}\mathrm{logaritham}\mathrm{\&}\mathrm{y}\mathrm{is}\mathrm{a}$$$$\mathrm{normal}\mathrm{number}.\mathrm{bt}\mathrm{you}\mathrm{mean}\mathrm{that}\mathrm{y}\mathrm{is}\mathrm{a}$$$$\mathrm{power}\mathrm{of}\mathrm{x}.$$

Commented by Shamim last updated on 03/Oct/19

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

Commented by Rio Michael last updated on 03/Oct/19

$$yourwelcomsir$$

Commented by MJS last updated on 03/Oct/19

$${\log }_{x}y=\frac{\log y}{\log x}=6$$$${\log }_{14x}8y=\frac{\log y+3\log 2}{\log x+\log 7+\log 2}=3$$$$$$$$\log y=6\log x$$$$\log y=3(\log x+\log 7)$$$$$$$$6\log x=3(\log x+\log 7)$$$$\log x=\log 7$$$$x=7$$$$\log y=6\log 7$$$$y=7^6$$

Commented by MJS last updated on 03/Oct/19

$$\left(1\right)z=x^y\iff y={\log }_{x}z$$$$\left(2\right)$$$$\log z=\log x^y$$$$\log z=y\log x\Rightarrow y=\frac{\log z}{\log x}$$$$$$$$\Rightarrow {\log }_{x}z=\frac{\log z}{\log x}$$

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

$$\mathrm{just}\mathrm{change}\mathrm{the}\mathrm{names}\mathrm{of}\mathrm{the}\mathrm{variables}$$$$z={\log }_{x}y\iff x^z=y$$$$\log x^z=\log y\Rightarrow z\log x=\log y\Rightarrow z=\frac{\log y}{\log x}$$$$$$$$\Rightarrow {\log }_{x}y=\frac{\log y}{\log x}$$

Commented by Shamim last updated on 03/Oct/19

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

Commented by Shamim last updated on 03/Oct/19

$$\mathrm{plz}\mathrm{give}\mathrm{me}\mathrm{the}\mathrm{law}---$$$$$$$${\log }_{\mathrm{x}}\mathrm{y}=\frac{\log \mathrm{y}}{\log \mathrm{x}}....\mathrm{plz}\mathrm{describe}\mathrm{details}.$$

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