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Question Number 193368 by MATHEMATICSAM last updated on 11/Jun/23

$$\mathrm{If}{\log }_{a}y=\frac{1}{3}\mathrm{and}{\log }_{8}a=x+1\mathrm{then}\mathrm{show}$$$$\mathrm{that}y=2^{x+1}$$

Answered by aba last updated on 11/Jun/23

$${\log }_8\mathrm{a}=\mathrm{x}+1\Rightarrow \ln \left(\mathrm{a}\right)=3(\mathrm{x}+1)\ln \left(2\right)\Rightarrow \ln \left(\mathrm{a}\right)=3\ln \left(2^{\mathrm{x}+1}\right)\left(1\right)$$$${\log }_{\mathrm{a}}\mathrm{y}=\frac{1}{3}\Rightarrow \ln \left(\mathrm{a}\right)=3\ln \left(\mathrm{y}\right)\left(2\right)$$$$\left(1\right)=\left(2\right)\Rightarrow \ln \left(\mathrm{y}\right)=\ln \left(2^{\mathrm{x}+1}\right)\Rightarrow \mathrm{y}=2^{\mathrm{x}+1}✓$$

Answered by aba last updated on 11/Jun/23

$${\log }_{\mathrm{a}}\left(\mathrm{y}\right)=\frac{1}{3}=\frac{{\log }_2\left(\mathrm{a}\right)}{{3\log }_2\left(\mathrm{a}\right)}\Rightarrow {\log }_{\mathrm{a}}\left(\mathrm{y}\right)=\frac{{\log }_8\left(\mathrm{a}\right)}{{\log }_2\left(\mathrm{a}\right)}$$$$\Rightarrow {\log }_{\mathrm{a}}\left(\mathrm{y}\right)=\frac{\mathrm{x}+1}{{\log }_2\left(\mathrm{a}\right)}\Rightarrow \ln \left(\mathrm{y}\right)=(\mathrm{x}+1)\ln \left(2\right)$$$$\Rightarrow \mathrm{y}=2^{\mathrm{x}+1}✓$$

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