Resolver calculo de logaritmo
23 jun 2013, 18:48
Alguém pode me ajudar a resolver esse calculo de logaritmo
Sendo \(log_{x}2=a , log_{x}3=b\) calcule \(log_{x}{\sqrt[3]{12}}\)
Obrigada
Sendo \(log_{x}2=a , log_{x}3=b\) calcule \(log_{x}{\sqrt[3]{12}}\)
Obrigada
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Re: Resolver calculo de logaritmo
24 jun 2013, 00:35
\(log_x\sqrt[3]{12}= log_x\ 12^{\frac{1}{3}}=\frac{1}{3}log_x(2.2.3)=\frac{1}{3}log_x 2+ \frac{1}{3}log_x 2 + \frac{1}{3}log_x 3= \frac{2}{3}log_x2+\frac{1}{3}log_x3= \frac{2}{3}a+\frac{1}{3}b\)
Re: Resolver calculo de logaritmo
24 jun 2013, 01:46
\(\log_x \sqrt[3]{12} =\)
\(\log_x 12^{\frac{1}{3}} =\)
\(\frac{1}{3} \cdot \log_x 12 =\)
\(\frac{1}{3} \cdot \log_x (2^2 \cdot 3) =\)
\(\frac{1}{3} \cdot \left ( \log_x 2^2 + \log_x 3 \right ) =\)
\(\frac{1}{3} \cdot \log_x 2^2 + \frac{1}{3} \cdot \log_x 3 =\)
\(\frac{2}{3} \cdot \log_x 2 + \frac{1}{3} \cdot \log_x 3 =\)
\(\frac{2}{3} \cdot a + \frac{1}{3} \cdot b =\)
\(\fbox{\frac{2a}{3} + \frac{b}{3}}\)
\(\log_x 12^{\frac{1}{3}} =\)
\(\frac{1}{3} \cdot \log_x 12 =\)
\(\frac{1}{3} \cdot \log_x (2^2 \cdot 3) =\)
\(\frac{1}{3} \cdot \left ( \log_x 2^2 + \log_x 3 \right ) =\)
\(\frac{1}{3} \cdot \log_x 2^2 + \frac{1}{3} \cdot \log_x 3 =\)
\(\frac{2}{3} \cdot \log_x 2 + \frac{1}{3} \cdot \log_x 3 =\)
\(\frac{2}{3} \cdot a + \frac{1}{3} \cdot b =\)
\(\fbox{\frac{2a}{3} + \frac{b}{3}}\)