Question

Find x, so that (2^-1+4^-1+6^-1+8^-1)^x=1

Answer 1

Answer:

$= {({2}^{ - 1} + {4}^{ - 1} + {6}^{ - 1} + {8}^{ - 1})}^{x} </p><p>\\ = {( \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} )}^{x} </p><p>\\ = {( \frac{192 + 96 + 64 + 48}{384} )}^{x} </p><p>\\ = {(\frac{400}{384})}^{x} </p><p>\\ = {(\frac{50}{48})}^{x} </p><p>= {(\frac{25}{24} )}^{x} </p><p>\\LHS = RHS</p><p>\\ {( \frac{25}{24} )}^{x} = 1 </p><p>\\ {( \frac{25}{24} )}^{x} </p><p>= {( \frac{25}{24} )}^{0} </p><p>\\ = x = 0$

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