25 nov 2012, 14:00
25 nov 2012, 23:15
25 nov 2012, 23:22
danjr5 Escreveu:Calculemos \(x\):
\(x = \log_2 2\sqrt{2}\)
\(x = \log_2 \sqrt{2^3}\)
\(x = \log_2 2^{\frac{3}{2}}\)
\(x = \frac{3}{2} \cdot \log_2 2\)
\(\fbox{x = \frac{3}{2}}\)
Agora, \(y\):
\(y = \log_{0,01} 10\)
\(y = \frac{\log_{10} 10}{\log_{10} 0,01}\)
\(y = \frac{1}{\log_{10} 10^{- 2}}\)
\(y = \frac{1}{- 2}\)
\(\fbox{y = - \frac{1}{2}}\)
Portanto,
\(x + y = \frac{3}{2} + \left ( - \frac{1}{2} \right )\)
\(x + y = \frac{3}{2} - \frac{1}{2}\)
\(\fbox{\fbox{\fbox{x + y = 1}}}\)