Question

Considerando a aproximação log2 5= 2,32, obtenha os valores de:
a) log(Base2) 10
b) log(Base2) 500
c) log(Base2) 1600
d) log(Base2) ³√¯ 0,2
e) log(Base2) (64/125)

Answer 1

a)
$\mathrm{\ell og_{2}\,}{10}\\ \\ =\mathrm{\ell og_{2}\,}{\left(2 \times 5 \right )}\\ \\ =\mathrm{\ell og_{2}\,}{2}+\mathrm{\ell og_{2}\,}{5}\\ \\ =1+2,32 \Rightarrow \boxed{\mathrm{\ell og_{2}\,}{10}=3,32}$


b)
$\mathrm{\ell og_{2}\,}{500}\\ \\ =\mathrm{\ell og_{2}\,}{\left(2^{2} \times 5^{3} \right )}\\ \\ =\mathrm{\ell og_{2}\,}\left(2^{2} \right )+\mathrm{\ell og_{2}\,}{\left(5^{3} \right )}\\ \\ =2\,\mathrm{\ell og_{2}\,}{2}+3\,\mathrm{\ell og_{2}\,}5\\ \\ =2 \times 1+3 \times 2,32 \Rightarrow \boxed{\mathrm{\ell og_{2}\,}{500}=8,96}$


c)
$\mathrm{\ell og_{2}\,}{1\,600}\\ \\ =\mathrm{\ell og_{2}\,}{\left(2^{6} \times 5^{2} \right )}\\ \\ =\mathrm{\ell og_{2}\,}{\left(2^{6} \right )}+\mathrm{\ell og_{2}\,}{\left(5^{2} \right )}\\ \\ =6\,\mathrm{\ell og_{2}\,}{2}+2\,\mathrm{\ell og_{2}\,}{5}\\ \\ =6 \times 1+2 \times 2,32 \Rightarrow \boxed{\mathrm{\ell og_{2}\,}{1\,600}=10,64}$


d)
$\mathrm{\ell og_{2}\,}{\sqrt[3]{0,2}}\\ \\ =\mathrm{\ell og_{2}\,}{\sqrt[3]{\frac{1}{5}}}\\ \\ =\mathrm{\ell og_{2}\,}{\left[ \left(\frac{1}{5} \right )^{1/3}\right ]}\\ \\ \\ =\frac{1}{3}\mathrm{\ell og_{2}\,}{\frac{1}{5}}\\ \\ =\frac{1}{3}\left(\mathrm{\ell og_{2}\,}{1}-\mathrm{\ell og_{2}\,}{5} \right )\\ \\ =\frac{0-2,32}{3} \Rightarrow \boxed{\mathrm{\ell og_{2}\,}{\sqrt[3]{0,2}}=-0,77}$


e)
$\mathrm{\ell og_{2}\,}{\left(\frac{64}{125} \right )}\\ \\ =\mathrm{\ell og_{2}\,}{\left(\frac{2^{6}}{5^{3}} \right )}\\ \\ =\mathrm{\ell og_{2}\,}{\left(2^{6} \right )}-\mathrm{\ell og_{2}\,}{\left(5^{3} \right )}\\ \\ =6\,\mathrm{\ell og_{2}\,}{2}-3\,\mathrm{\ell og_{2}\,}{5}\\ \\ =6 \times 1-3 \times 2,32 \Rightarrow \boxed{\mathrm{\ell og_{2}\,}{\left(\frac{64}{125} \right )}=-0,96}$