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Question Number 170427 by mathlove last updated on 23/May/22

$${14}^b=7^{b-a}$$$${49}^{\frac{a}{b}}=?$$

Answered by thfchristopher last updated on 23/May/22

$${14}^b=7^{b-a}$$$$\Rightarrow \left(2^b\right)\left(7^b\right)=7^{b-a}$$$$\Rightarrow {\log }_7\left[\left(2^b\right)\left(7^b\right)\right]={\log }_7{7}^{b-a}$$$$\Rightarrow b{\log }_{7}2+b{\log }_{7}7=b{\log }_{7}7-a{\log }_{7}7$$$$\Rightarrow b{\log }_{7}2=-a$$$$\Rightarrow \frac{a}{b}=-{\log }_{7}2$$$$\therefore {49}^{\frac{a}{b}}$$$$={\left(7^2\right)}^{-{\log }_{7}2}$$$$=\frac{1}{7^{{\log }_{7}4}}$$$$=\frac{1}{4}$$

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