Logaritmos |
Actividad (#1)Calcula:
$$
\left.\begin{array}{l}
a) \displaystyle{\log_{2}(16)} &
b) \displaystyle{\log_{2}(0'125)} &
c) \displaystyle{\log_{6}(\frac{1}{216})} \\
d) \displaystyle{\log_{5}(0'04)} &
e) \displaystyle{\ln(e^{12})} &
f) \displaystyle{\ln(e^{-\frac{1}{4}})} \\
\end{array}\right.
$$
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Actividad (#2)Sabiendo que \( \left.\begin{array}{lll} \displaystyle{\log_{2}(M)=3'5} & \hbox{y} & \displaystyle{\log_{2}(N)=-1'4,} & \hbox{determina:}\\ \end{array}\right. \)
$$
\left.\begin{array}{l}
a) \displaystyle{\log_{2}(\frac{M \cdot N}{4})} &
b) \displaystyle{\log_{2}(\frac{2\sqrt{M}}{N^3})} \\
\end{array}\right.
$$
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Actividad (#3)Expresa como un solo logaritmo:
$$
\left.\begin{array}{l}
a) \displaystyle{\ln(M)+2\ln(N)-\ln(P)} \\
\end{array}\right.
$$
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Actividad (#4)Determina la relación que existe entre \(x\) e \(y\) para que se cumpla:
$$
\left.\begin{array}{l}
a) \displaystyle{\ln(y)=x+\ln(7)} &
b) \displaystyle{\ln(y)=2x-\ln(5)}\\
\end{array}\right.
$$
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Actividad (#5)Si consideramos que \( \left.\begin{array}{lll} \displaystyle{\log_{}(3)=0'477} & \hbox{y} & \displaystyle{\log_{}(2)=0'301,} & \hbox{determina:}\\ \end{array}\right. \)
$$
\left.\begin{array}{l}
a) \displaystyle{\log(3000)} &
b) \displaystyle{\log(50)} &
c) \displaystyle{\log(\sqrt[5]{60})} &
d) \displaystyle{\log(\sqrt{\frac{9}{8}})} &
e) \displaystyle{\log(0'0012)} \\
\end{array}\right.
$$
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