Question

if log x base 2 + log x base 4 + log x base 16 = 22/4. find x

Answer 1

So answer is apprximated as 8

Answer 2

$log_2\ x+log_4\ x+log_{16}\ x= \frac{22}{4} \\ \\log_2\ x+log_{2^2}\ x+log_{2^4}\ x= \frac{22}{4} \\\\log_2\ x+ \frac{1}{2} log_{2}\ x+ \frac{1}{4} log_{2}\ x= \frac{22}{4} \\\\log_{2}\ x \times ( 1 + \frac{1}{2} + \frac{1}{4} )= \frac{22}{4} \\\\log_{2}\ x \times ( \frac{4+2+1}{4})= \frac{22}{4} \\\\log_{2}\ x \times ( \frac{7}{4})= \frac{22}{4} \\ \\ log_{2}\ x= \frac{22}{4} \times \frac{4}{7} = \frac{22}{7} \\ \\ x=2^{ \frac{22}{7} }$