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Question Number 97390 by shaxzod last updated on 07/Jun/20

$$compare{log}_{2}3with{log}_{3}4$$

Answered by mr W last updated on 07/Jun/20

$$9>8$$$$3^2>2^3$$$$2{\log }_{2}3>3$$$$\Rightarrow {\log }_{2}3>\frac{3}{2}=\frac{9}{6}$$$$$$$$8<9$$$$2^3=4^{\frac{3}{2}}<3^2$$$$\frac{3}{2}{\log }_{3}4<2$$$$\Rightarrow {\log }_{3}4<\frac{4}{3}=\frac{8}{6}<\frac{9}{6}=\frac{3}{2}<{\log }_{2}3$$$$i.e.{\log }_{2}3>{\log }_{3}4$$

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