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x^(log(y)-log(z))*y^(log(z)-log(x))*z^(log(x)-log(y))=1

 Mar 18, 2016

Best Answer 

 #1
avatar+26376 
+10

x^(log(y)-log(z))*y^(log(z)-log(x))*z^(log(x)-log(y))=1

 

\(\small{ \begin{array}{rcll} x^{ \log{(y)}-\log{(z)} } \cdot y^{ \log{(z)}-\log{(x)} } \cdot z^{ \log{(x)}-\log{(y)}} &\overset{?}{=} & 1 \quad | \quad \log{()} \\\\ \log{( x^{ \log{(y)}-\log{(z)} } \cdot y^{ \log{(z)}-\log{(x)} } \cdot z^{ \log{(x)}-\log{(y)}} )} &\overset{?}{=} & \log{(1)} \quad | \quad \log{(1)} = 0 \\\\ \log{( x^{ \log{(y)}-\log{(z)} } )} + \log{( y^{ \log{(z)}-\log{(x)} } )} + \log{( z^{ \log{(x)}-\log{(y)}} )} &\overset{?}{=} & 0 \\\\ \end{array} } \)

\(\small{ \begin{array}{rcll} [~ \log{(y)}-\log{(z)} ~] \cdot \log{(x)} \\ +[~ \log{(z)}-\log{(x)} ~] \cdot \log{(y)} \\ +[~ \log{(x)}-\log{(y)} ~] \cdot \log{(z)} &\overset{?}{=} & 0 \end{array} } \)

 

\(\small{ \begin{array}{rcll} \log{(y)} \cdot \log{(x)} - \log{(z)}\cdot \log{(x)} \\ + \log{(z)} \cdot \log{(y)} - \log{(x)} \cdot \log{(y)} \\ + \log{(x)} \cdot \log{(z)} - \log{(y)} \cdot \log{(z)} &\overset{?}{=} & 0 \end{array} }\)

 

\(\small{ \begin{array}{rcll} \log{(y)} \cdot \log{(x)}- \log{(x)} \cdot \log{(y)}\\ - \log{(z)}\cdot \log{(x)} + \log{(x)} \cdot \log{(z)}\\ + \log{(z)} \cdot \log{(y)} - \log{(y)} \cdot \log{(z)} &\overset{?}{=} & 0 \\\\ 0+0+0 &\overset{?}{=} & 0 \\\\ \mathbf{0} & \mathbf{=} & \mathbf{0} \end{array} }\)

 

laugh

 Mar 18, 2016
 #1
avatar+26376 
+10
Best Answer

x^(log(y)-log(z))*y^(log(z)-log(x))*z^(log(x)-log(y))=1

 

\(\small{ \begin{array}{rcll} x^{ \log{(y)}-\log{(z)} } \cdot y^{ \log{(z)}-\log{(x)} } \cdot z^{ \log{(x)}-\log{(y)}} &\overset{?}{=} & 1 \quad | \quad \log{()} \\\\ \log{( x^{ \log{(y)}-\log{(z)} } \cdot y^{ \log{(z)}-\log{(x)} } \cdot z^{ \log{(x)}-\log{(y)}} )} &\overset{?}{=} & \log{(1)} \quad | \quad \log{(1)} = 0 \\\\ \log{( x^{ \log{(y)}-\log{(z)} } )} + \log{( y^{ \log{(z)}-\log{(x)} } )} + \log{( z^{ \log{(x)}-\log{(y)}} )} &\overset{?}{=} & 0 \\\\ \end{array} } \)

\(\small{ \begin{array}{rcll} [~ \log{(y)}-\log{(z)} ~] \cdot \log{(x)} \\ +[~ \log{(z)}-\log{(x)} ~] \cdot \log{(y)} \\ +[~ \log{(x)}-\log{(y)} ~] \cdot \log{(z)} &\overset{?}{=} & 0 \end{array} } \)

 

\(\small{ \begin{array}{rcll} \log{(y)} \cdot \log{(x)} - \log{(z)}\cdot \log{(x)} \\ + \log{(z)} \cdot \log{(y)} - \log{(x)} \cdot \log{(y)} \\ + \log{(x)} \cdot \log{(z)} - \log{(y)} \cdot \log{(z)} &\overset{?}{=} & 0 \end{array} }\)

 

\(\small{ \begin{array}{rcll} \log{(y)} \cdot \log{(x)}- \log{(x)} \cdot \log{(y)}\\ - \log{(z)}\cdot \log{(x)} + \log{(x)} \cdot \log{(z)}\\ + \log{(z)} \cdot \log{(y)} - \log{(y)} \cdot \log{(z)} &\overset{?}{=} & 0 \\\\ 0+0+0 &\overset{?}{=} & 0 \\\\ \mathbf{0} & \mathbf{=} & \mathbf{0} \end{array} }\)

 

laugh

heureka Mar 18, 2016
 #2
avatar
+4

x^(log(y)-log(z))*y^(log(z)-log(x))*z^(log(x)-log(y))=1

 

x^((log(y))/(log(10))-(log(z))/(log(10))) y^((log(z))/(log(10))-(log(x))/(log(10))) z^((log(x))/(log(10))-(log(y))/(log(10))) = 1. THE STATEMENT IS TUE!!.

 Mar 18, 2016

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