Question

X =1+log bc base a , Y = 1+ log ca base b and Z = 1+ log ab base c then prove that xy + yz + zx= xyz​

Answer 1

$\red{ \huge{ \underline{ \overline{ \mid{ \bold{ \purple{ \: SOLUTION \: \: \red{ \mid}}}}}}}}$

$\underline{ \underline{ \bold{\: \: GIVEN \: \: }}} \to \\ \\ \bold{x = 1 + log_{a}bc} \\ \\ \bold{y = 1 + log_{b}ca} \\ \\ \bold{z = 1 + log_{c}ab } \\ \\ \\ \underline{ \underline{ \bold{ \: \: TO \: \: PROVE \: \: }}} \to \: \: \: \: \bold{xy + yz + zx = xyz}$

$\bold{Now,} \\ \\ \bold{x = 1 + log_{a}bc} \\ \\ \bold{ \Longrightarrow \: x = log_{a}a + log_{a}bc } \\ \\ \bold{ \Longrightarrow \: x = log_{a}abc} \\ \\ \bold{ \Longrightarrow \: \frac{1}{x} = \frac{1}{ log_{a}abc} } \\ \\ \bold{ \Longrightarrow \: \frac{1}{x} = log_{abc}a \: \: - - - - > (1) } \\ \\ \\ \bold{Similarly,} \\ \\ \bold{ \frac{1}{y} = log_{abc}b \: \: - - - - > (2)} \\ \\ \bold{And,} \\ \\ \bold{\frac{1}{z} = log_{abc}c \: \: - - - - > (3)}$

$\bold{ \therefore \: \: (1) + (2) + (3) \: \: \Longrightarrow \: } \\ \\ \bold{\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = log_{abc}a + log_{abc} + log_{abc}c} \\ \\ \bold{ \Longrightarrow \frac{yz + zx + xy}{xyz} = log_{abc}(abc)} \\ \\ \bold{ \Longrightarrow \frac{xy + yz + zx}{xyz} = 1} \\ \\ \bold{ \Longrightarrow \: \underline{ \: \: xy + yz + zx = xyz \: \: }}$

$\bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \underline{ \: \: Hence \: \: Proved. \: \: }}}$

Answer 2

$\huge\mathtt\blue{Answer}$

$\begin{lgathered}\underline{ \underline{ \bold{\: \: GIVEN \: \: }}} \to \\ \\ \bold{x = 1 + log_{a}bc} \\ \\ \bold{y = 1 + log_{b}ca} \\ \\ \bold{z = 1 + log_{c}ab } \\ \\ \\ \underline{ \underline{ \bold{ \: \: TO \: \: PROVE \: \: }}} \to \: \: \: \: \bold{xy + yz + zx = xyz}\end{lgathered}$

$\begin{lgathered}\bold{Now,} \\ \\ \bold{x = 1 + log_{a}bc} \\ \\ \bold{ \Longrightarrow \: x = log_{a}a + log_{a}bc } \\ \\ \bold{ \Longrightarrow \: x = log_{a}abc} \\ \\ \bold{ \Longrightarrow \: \frac{1}{x} = \frac{1}{ log_{a}abc} } \\ \\ \bold{ \Longrightarrow \: \frac{1}{x} = log_{abc}a \: \: - - - - > (1) } \\ \\ \\ \bold{Similarly,} \\ \\ \bold{ \frac{1}{y} = log_{abc}b \: \: - - - - > (2)} \\ \\ \bold{And,} \\ \\ \bold{\frac{1}{z} = log_{abc}c \: \: - - - - > (3)}\end{lgathered}$

$\begin{lgathered}\bold{ \therefore \: \: (1) + (2) + (3) \: \: \Longrightarrow \: } \\ \\ \bold{\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = log_{abc}a + log_{abc} + log_{abc}c} \\ \\ \bold{ \Longrightarrow \frac{yz + zx + xy}{xyz} = log_{abc}(abc)} \\ \\ \bold{ \Longrightarrow \frac{xy + yz + zx}{xyz} = 1} \\ \\ \bold{ \Longrightarrow \: \underline{ \: \: xy + yz + zx = xyz \: \: }}\end{lgathered}$

Hence Proved.