Question

2^x=3^y=36^z, then prove that z-¹=2(x-¹+y-¹)​

Answer 1

Step-by-step explanation:

$Let \: {2}^{x} = {3}^{y} = {6}^{2z \: } = k \\ then \\ 2 = {k}^{ \frac{1}{x} } \\ 3 = {k}^{ \frac{1}{y} } \\ 6 = {k}^{ \frac{1}{2z} } \\ we \: know \: that \: \\ 2 \times 3 = 6 \: \: \: \: \: \: .....(1)\\ {x}^{a} \times {x}^{b} = {x}^{a + b} \: \: \: \: \: ....(2) \\ \:now \: put \: the \: values \: of \: 2 \: \: \: 3 \: \: and \: 6 \\ {k}^{ \frac{1}{x} } \times {k}^{ \frac{1}{y} } = {k}^{ \frac{1}{2z} } \\ \frac{1}{x} + \frac{1}{y} = \frac{1}{2z} \\ {x}^{ - 1 } + {y}^{ - 1} = \frac{ {z}^{ - 1} }{2} \\2( {x}^{ - 1 } + {y}^{ - 1} )= {z}^{ - 1}$