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Question Number 20245 by virus last updated on 24/Aug/17

$$\mathrm{y}=2^{\frac{1}{{log}_{x}8}}$$$$thenx=?$$

Answered by Tinkutara last updated on 24/Aug/17

$$y=2^{{\log }_{8}x}=2^{\frac{1}{3}{\log }_{2}x}$$$$y^3=2^{{\log }_{2}x}=x$$$$\Rightarrow x=y^3$$

Answered by Rauny last updated on 27/Jan/19

$$\frac{1}{\left(\frac{ln8}{lnx}\right)}={log}_{2}y$$$$\frac{lnx}{ln8}=\frac{lny}{ln2}$$$$lnx=\frac{ln8}{ln2}\cdot lny={log}_{2}8\cdot lny$$$$=3lny$$$$x=e^{3lny}$$$$=y^3$$$$\therefore x=y^3$$

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