Question

Sabendo que log2=0,3 e log3 =0,48, o log 6000 na base 2?

Answer 1

log 2 = 0,3
log 3 = 0,48

log 6000 = log 6.1000 = log 6 + log 1000
log 6 = log 2.3 = log 2 + log 3
log 1000 = log 10³ = 3 log 10

log 6000 = log 2 + log 3 + 3log 10
log 6000 = 0,3 + 0,48 + 3.1
log 6000 = 0,3 + 0,48 + 3
log 6000 = 3,78

Espero ter ajudado.
 

Answer 2

Utilizando a lei de mudança de base (para a base $10$ neste caso), temos que

$\mathrm{\ell og}_{2}\,6\,000=\dfrac{\mathrm{\ell og}\,6\,000}{\mathrm{\ell og\,}2}\\ \\ \\ \mathrm{\ell og}_{2}\,6\,000=\dfrac{\mathrm{\ell og}\,(2\cdot 3\cdot 10^{3})}{\mathrm{\ell og\,}2}\\ \\ \\ \mathrm{\ell og}_{2}\,6\,000=\dfrac{\mathrm{\ell og}\,2+\mathrm{\ell og\,}3+\mathrm{\ell og\,}(10)^{3}}{\mathrm{\ell og\,}2}\\ \\ \\ \mathrm{\ell og}_{2}\,6\,000=\dfrac{\mathrm{\ell og}\,2+\mathrm{\ell og\,}3+3\mathrm{\,\ell og\,}10}{\mathrm{\ell og\,}2}\\ \\ \\ \mathrm{\ell og}_{2}\,6\,000=\dfrac{\mathrm{\ell og}\,2+\mathrm{\ell og\,}3+3\cdot 1}{\mathrm{\ell og\,}2}\\ \\ \\ \mathrm{\ell og}_{2}\,6\,000=\dfrac{\mathrm{\ell og}\,2+\mathrm{\ell og\,}3+3}{\mathrm{\ell og\,}2}$


Substituindo os valores dados, temos

$\mathrm{\ell og}_{2}\,6\,000=\dfrac{0,3+0,48+3}{0,3}\\ \\ \\ \mathrm{\ell og}_{2}\,6\,000=\dfrac{3,78}{0,3}\\ \\ \\ \boxed{\begin{array}{c}\mathrm{\ell og}_{2}\,6\,000=12,6\end{array}}$