Question

Find the value of 49 ^{(1 - 1og} ₇²⁾ + 5 ^-log₅^4.

Answer 1

$$\begin{gathered} {49}^{\left(1-{\log }_{7}2\right)}+5^{-{\log }_{5}4} \\ =7^{2\left({\log }_{7}7-{\log }_{7}2\right)}+5^{{\log }_5{4}^{-1}} \\ =7^{2\left({\log }_7\frac{7}{2}\right)}+5^{{\log }_5\frac{1}{4}} \\ =7^{{\log }_7{\left(\frac{7}{2}\right)}^2}+5^{{\log }_5\frac{1}{4}} \\ Nowu\sin g{e}^{{\log }_{e}x}=x \\ =\frac{49}{4}+\frac{1}{4}=\frac{25}{2} \end{gathered}$$