Question

find the value of x
(2^-1+4^-1+6^-1+8^-1)^x=1​

Answer 1

Answer:

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Answer 2

ANSWER:-

The required value of x is 0

Solution :-

Given :-

$( {2}^{ - 1} + {4}^{ - 1} + {6}^{ - 1} + {8}^{ - 1} )^ {x} = 1$

Next,

$apply \: {x}^{ - m} = \frac{1}{ {x}^{m} } \: in \: above \: equation \\ = (\frac{1}{ {2}^{1} } + \frac{1}{ {4}^{1} } + \frac{1}{ {6}^{1} } + \frac{1}{ {8}^{1} } )^{x} = 1 \\ = ( \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} )^{x} = 1$

Now,

We have to make denominator of equation (1) equal -

$( \frac{1}{2} \times \frac{12}{12} + \frac{1}{4} \times \frac{6}{6} + \frac{1}{6} \times \frac{4}{4} + \frac{1}{8} \times \frac{3}{3} )^{x} = 1 \\ \\ = ( \frac{12 + 6 + 4 + 3}{24}) ^{x} = 1 \\ \\ = (\frac{25}{24} )^{x} = 1$

This can only be possible when x = 0

$as \: a ^{0} = 1 \: {exponent \: law} \\ \\ = > ( \frac{25}{24} )^{0} = 1 \\ \\ => 1 = 1$

Therefore, LHS = RHS

Hence the value of x = 0.

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