Question

Log2, x + Log4 , x + log8, x +log.16, x = -6,25

Answer 1

Primeiro, colocamos todos na base 2:

$log_{2} x + \frac{ log_{2}x}{ log_{2}4 } + \frac{ log_{2}x}{ log_{2}8} + \frac{ log_{2}x}{ log_{2}16} = -625. 10^{-2} \\ log_{2}x+ \frac{1}{2} log_{2}x + \frac{1}{3} log_{2}x + \frac{1}{4}log_{2}x = -625. 10^{-2} \\ \frac{12log_{2}x + 6log_{2}x +4log_{2}x+3log_{2}x}{12} = -625. 10^{-2} \\ \frac{25log_{2}x}{12} = -625. 10^{-2} \\ \frac{log_{2}x}{12} = -25. 10^{-2} \\ log_{2}x = -300. 10^{-2} \\ log_{2}x= -3 \\ 2^{-3} =x \\ x= \frac{1}{8}$

Espero ter ajudado :)