Question

Quero que me ajudem é para amanhã
Log4 1/256 - Log 1/16 2 rapido por favor me ajudem

Answer 1

$\log_4\left(\dfrac{1}{256}\right)=x\\\\4^{x}=\dfrac{1}{256}\\\\(2^2)^x=\dfrac{1}{2^8}\\\\2^{2x}=2^{-8}\\\\2x=-8\\\\x=-4\\\\\\\log_{\frac{1}{16}}(2)=x\\\\\left(\dfrac{1}{16}\right)^x=2\\\\\left(\dfrac{1}{2^4}\right)^x=2^1\\\\(2^{-4})^x=2^1\\\\-4x=1\\\\x=-\dfrac14$

Agora que temos o valor dos dois logs calculemos a expressao:

$-4-\left(-\dfrac14\right)=-4+\dfrac14=-\dfrac{15}{4}$