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

Given that 1+log_3 x =log_(27) y, express y  in terms of x.

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{1}+\mathrm{log}_{\mathrm{3}} \mathrm{x}\:=\mathrm{log}_{\mathrm{27}} \mathrm{y},\:\mathrm{express}\:\mathrm{y} \\ $$$$\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{x}. \\ $$

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

log_3 3+log_3 x=log_3^3  y⇒log_3 3x=(1/3)log_3 y=log_3 y^(1/3) ⇒  3x=y^(1/3) ⇒y=(3x)^3

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

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

1=log_(27) y−log_(3  ) x  1=(1/3)log _(3 ) y−log _(3 ) x  3=log _3 y−3log _3 x  3=log_3  ((y/x^3 ))  27=(y/x^3 )  27x^3 =y    ......

$$\mathrm{1}={log}_{\mathrm{27}} {y}−{log}_{\mathrm{3}\:\:} {x} \\ $$$$\mathrm{1}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{log}\:_{\mathrm{3}\:} {y}−\mathrm{log}\:_{\mathrm{3}\:} {x} \\ $$$$\mathrm{3}=\mathrm{log}\:_{\mathrm{3}} {y}−\mathrm{3log}\:_{\mathrm{3}} {x} \\ $$$$\mathrm{3}=\mathrm{log}_{\mathrm{3}} \:\left(\frac{{y}}{{x}^{\mathrm{3}} }\right) \\ $$$$\mathrm{27}=\frac{{y}}{{x}^{\mathrm{3}} } \\ $$$$\mathrm{27}{x}^{\mathrm{3}} ={y} \\ $$$$ \\ $$$$...... \\ $$

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