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Question Number 53675 by pieroo last updated on 24/Jan/19

$$\mathrm{Given}\mathrm{that}1+{\log }_3\mathrm{x}={\log }_{27}\mathrm{y},\mathrm{express}\mathrm{y}$$$$\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{x}.$$

Answered by kaivan.ahmadi last updated on 24/Jan/19

$${\log }_{3}3+{\log }_3\mathrm{x}={\log }_{3^3}\mathrm{y}\Rightarrow {\log }_{3}3\mathrm{x}=\frac{1}{3}{\log }_3\mathrm{y}={\log }_3{\mathrm{y}}^{\frac{1}{3}}\Rightarrow$$$$3\mathrm{x}={\mathrm{y}}^{\frac{1}{3}}\Rightarrow \mathrm{y}={\left(3\mathrm{x}\right)}^3$$

Answered by peter frank last updated on 24/Jan/19

$$1={log}_{27}y-{log}_{3}x$$$$1=\frac{1}{3}\log {}_{3}y-\log {}_{3}x$$$$3=\log {}_{3}y-3\log {}_{3}x$$$$3={\log }_3\left(\frac{y}{x^3}\right)$$$$27=\frac{y}{x^3}$$$$27x^3=y$$$$$$$$......$$

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