Solve the following equations $-2={{\log }_{7}}x-3{{\log }_{x}}7$ [3] $\frac{{{2}^{2x}}}{21}=\frac{{

Question

Solve the following equations $-2={{\log }_{7}}x-3{{\log }_{x}}7$ [3] $\frac{{{2}^{2x}}}{21}=\frac{{{5}^{x}}}{6}$ [2]

Accepted Answer

(a) $x = \frac{1}{343}$ \text{ or } $x = 7$ (b) $x = -5.61$

2020 HYSS AM S4 P1 Q4