Question

if 2 to the power X = 3 to the power Y = 6 to the power - Z. Then prove 1/X + 1/Y + 1/Z = 0

Answer 1

${2}^{x} = {3}^{y} = {6}^{ - z} = k \\ 2 = {k}^{ \frac{1}{x} } \\ 3 = {k}^{ \frac{1}{y} } \\ 6 = {k}^{ \frac{1}{ - z} }$
Now we know that
$2 \times 3 = 6 \\ {k}^{ \frac{1}{x} } \times {k}^{ \frac{1}{y} } = {k}^{ \frac{1}{ - z} } \\ {k}^{ \frac{1}{x} + \frac{1}{y} } = {k}^{ \frac{1}{ - z} } \\ \frac{1}{x} + \frac{1}{y} = - \frac{1}{z} \\ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$
Mark it as the brainliest answer