If $x = {\log _a}(bc),y = {\log _b}(ca),z = {\log _c}(ab),$then which of the following is equal to $1$
If $x = {\log _a}(bc),y = {\log _b}(ca),z = {\log _c}(ab),$then which of the following is equal to $1$
- A
$x + y + z$
- B
${(1 + x)^{ - 1}} + {(1 + y)^{ - 1}} + {(1 + z)^{ - 1}}$
- C
$xyz$
- D
None of these
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Solution set of equation
$\left| {1 - {{\log }_{\frac{1}{6}}}x} \right| + \left| {{{\log }_2}x} \right| + 2 = \left| {3 - {{\log }_{\frac{1}{6}}}x + {{\log }_{\frac{1}{2}}}x} \right|$ is $\left[ {\frac{a}{b},a} \right],a,b, \in N,$ then the value of $(a + b)$ is
Solution set of equation
$\left| {1 - {{\log }_{\frac{1}{6}}}x} \right| + \left| {{{\log }_2}x} \right| + 2 = \left| {3 - {{\log }_{\frac{1}{6}}}x + {{\log }_{\frac{1}{2}}}x} \right|$ is $\left[ {\frac{a}{b},a} \right],a,b, \in N,$ then the value of $(a + b)$ is