Alg9.4.30

Solve $\log_{16} x+\log_4 x+\log_2 x=7\,$

Readjusting the logs,

$\frac{\log x}{\log 16}+\frac{\log x}{\log 4}+\frac{\log x}{\log 2}=7\,$

Simplifying the denominators

$\frac{\log x}{4\log 2}+\frac{\log x}{2\log 2}+\frac{\log x}{\log 2}=7\,$

Simplifying further,

$\frac{\log x+2\log x+4\log x}{4\log 2}=7\,$

$\log x+2\log x+4\log x=28\log 2\,$

$\log (x\cdot x^2\cdot x^4)=\log 2^{28}\,$

Taking out logs on both sides,we get

$x^7=2^{28}\,$

$x=(2^{28})^{\frac{1}{7}}\,$

Hence the value of x is 16