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# What is the value of Log_d (abc)?

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If a, b, c and d are positive real numbers such that log_a  b= 8/9, log_b  c= -(3/4), log_c d = 2,

find the value of log_d (abc)

Feb 22, 2019

#1
+21848
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If a, b, c and d are positive real numbers such that
log_a  b= 8/9,
log_b  c= -(3/4),
log_c d = 2,
find the value of log_d (abc)

$$\begin{array}{|rclcl|} \hline \log_a(b) &=& \dfrac{\log_d(b)}{\log_d(a)} &=& \dfrac{\log_b(b)}{\log_b(a)} \\\\ & & \dfrac{\log_d(b)}{\log_d(a)} &=& \dfrac{\log_b(b)}{\log_b(a)} \quad | \quad \log_b(b) = 1 \\\\ & & \dfrac{\log_d(b)}{\log_d(a)} &=& \dfrac{1}{\log_b(a)} \quad | \quad \log_a(b)\log_b(a)=1 \\\\ & & \dfrac{\log_d(b)}{\log_d(a)} &=& \log_a(b) \\\\ & & \dfrac{\log_d(a)} {\log_d(b)}&=& \dfrac{1}{\log_a(b)} \\\\ & & \mathbf{\log_d(a)} & \mathbf{=} & \mathbf{ \dfrac{\log_d(b)}{\log_a(b)} } \qquad (1) \\ \hline \log_b(c) &=& \dfrac{\log_d(c)}{\log_d(b)} &=& \dfrac{\log_c(c)}{\log_c(b)} \\\\ & & \dfrac{\log_d(c)}{\log_d(b)} &=& \dfrac{\log_c(c)}{\log_c(b)} \quad | \quad \log_c(c) = 1 \\\\ & & \dfrac{\log_d(c)}{\log_d(b)} &=& \dfrac{1}{\log_c(b)} \quad | \quad \log_b(c)\log_c(b)=1 \\\\ & & \dfrac{\log_d(c)}{\log_d(b)} &=& \log_b(c) \\\\ & & \dfrac{\log_d(b)}{\log_d(c)} &=& \dfrac{1}{\log_b(c)} \\\\ & & \mathbf{\log_d(b)} &\mathbf{=}& \mathbf{\dfrac{\log_d(c)}{\log_b(c)}} \qquad (2) \\ \hline && \log_c(d) &=& \dfrac{\log_d(d)}{\log_d(c)} \quad | \quad \log_d(d) = 1 \\\\ && \log_c(d) &=& \dfrac{1}{\log_d(c)} \\\\ & & \mathbf{\log_d(c)} &\mathbf{=}& \mathbf{\dfrac{1}{\log_c(d)}} \qquad (3) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \log_d(abc) &=& \log_d(a)+\log_d(b)+\log_d(c) \quad | \quad \mathbf{\log_d(a)=\dfrac{\log_d(b)}{\log_a(b)} } \qquad (1) \\ &=& \dfrac{\log_d(b)}{\log_a(b)}+\log_d(b)+\log_d(c) \\ &=& \log_d(b) \left( \dfrac{1}{\log_a(b)}+1 \right) +\log_d(c) \quad | \quad \mathbf{\log_d(b)=\dfrac{\log_d(c)}{\log_b(c)}} \qquad (2) \\ &=& \dfrac{\log_d(c)}{\log_b(c)} \left( \dfrac{1}{\log_a(b)}+1 \right) +\log_d(c) \\ &=&\log_d(c)\left( \dfrac{1}{\log_b(c)} \left( \dfrac{1}{\log_a(b)}+1 \right) + 1 \right) \quad | \quad \mathbf{\log_d(c)=\dfrac{1}{\log_c(d)}} \qquad (3) \\ \mathbf{\log_d(abc)} & \mathbf{=} & \mathbf{\dfrac{1}{\log_c(d)}\left( \dfrac{1}{\log_b(c)} \left( \dfrac{1}{\log_a(b)}+1 \right) + 1 \right)} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{\log_d(abc)} & \mathbf{=} & \mathbf{\dfrac{1}{\log_c(d)}\left( \dfrac{1}{\log_b(c)} \left( \dfrac{1}{\log_a(b)}+1 \right) + 1 \right)} \\\\ & = & \dfrac{1}{2}\left( \dfrac{1}{ -\dfrac{3}{4} } \left( \dfrac{1}{\dfrac{8}{9}}+1 \right) + 1 \right) \\\\ & = & \dfrac{1}{2}\left( \dfrac{-4}{3} \Big( \dfrac{1}{8}+1 \Big) + 1 \right) \\\\ & = & \dfrac{1}{2}\left( \dfrac{-4}{3} \left( \dfrac{17}{8} \right) + 1 \right) \\\\ & = & \dfrac{1}{2}\left( \dfrac{-17}{6} + 1 \right) \\\\ & = & \dfrac{1}{2}\left( \dfrac{-11}{6} \right) \\\\ & \mathbf{=} & -\mathbf{\dfrac{11}{12}} \\ \hline \end{array}$$

Feb 22, 2019
edited by heureka  Feb 22, 2019