Question

Ao se fazer a mudança da base 2 para a base 10 e simplificar a expressão log2(2 raiz b, considerando log2= 0,301, chega-se a, aproximadamente

Answer 1

$\mathsf{Ola\´~Beyonce}$

$\mathsf{Fazendo~a~mudanca~de~bases}$
$\mathsf{\log_2(2\sqrt{b})=\Large\frac{log(2\sqrt{b})}{log(2)}}\\\\\mathsf{\log2\approx0,301}\\\\\mathsf{\log_2(2.b^\frac{1}{2})}=\frac{\log(2.b^\frac{1}{2})}{0,301}}\\\\\mathsf{\log_2(2)+\frac{1}{2}.\log_2(b)}=\frac{\log(2)+\frac{1}{2}.\log(b)}{0,301}\\\\\mathsf{\Big(1+\frac{1}{2}.\log_2(b)\Big).\Big(0,301\Big)={0,301+\frac{1}{2}.log(b)}}\\\\\mathsf{0,301+\frac{0,301}{2}.log_2(b)=0,301+\frac{1}{2}.\log(b)}\\\\\mathsf{\not{2}.\frac{0,301}{\not{2}}.\log_2(b)=\log(b)}\\\\\\\mathsf{\Large\boxed{\log_2(b)=\frac{\log(b)}{0,301}}}}}}\\\\\mathsf{ou}$

$\mathsf{\Large\boxed{\log_2(b)=\frac{\log(b)}{log(2)}}}\\\\\\\\\mathsf{Du\´vidas?~comente}$

Answer 2

Resposta:

resposta correta