Alg9.4.47

Solve $\log_2 x+\log_4 x+\log_{16} x=\frac{21}{4}\,$

$\frac{\log x}{\log 2}+\frac{\log x}{\log 4}+\frac{\log x}{\log 16}=\frac{21}{4}\,$

$\frac{\log x}{\log 2}+\frac{\log x}{2\log 2}+\frac{\log x}{4\log 2}=\frac{21}{4}\,$

$\frac{4\log x+2\log x+\log x}{4\log 2}=\frac{21}{4}\cdot4\log 2\,$

$28\log x=84\log 2\,$

$\log (x^{28})=\log (2^{84})\,$

Hence

$x=2^3\,$

Hence the value of x=8