Question

7) Find the value of x so that (2^-1 + 4^-1 + 6^-1 + 8^-1)^x = 1​

Answer 1

Answer:

the value of x=0

Step-by-step explanation:

Answer 2

Question

$\bf \:Find \: the \: value \: of \: x \: so \: that \:$

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

Answer

Identity used :-

$\bf \: {x}^{ - 1} = \dfrac{1}{x}$

$\bf \: {x}^{0} = 1$

Solution:-

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

$\bf\implies \: {(\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{8}) }^{x} = 1$

$\bf\implies \: {(\dfrac{12 + 6 + 4 + 3}{24}) }^{x} = 1$

$\bf\implies \: {(\dfrac{25}{24} )}^{x} = 1$

$\bf\implies \: {(\dfrac{25}{24}) }^{x} = {(\dfrac{25}{24})}^{0}$

$\bf\implies \:x = 0$

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Additional Information

Properties of exponents:-

$\bf \:{x}^{m} \times {x}^{n} = {x}^{m + n}$

$\bf \: {x}^{m} \div {x}^{n} = {x}^{m - n}$

$\bf \: {x}^{ - m} = \dfrac{1}{ {x}^{m} }$

$\bf \: {x}^{m} \times {y}^{m} = {(xy)}^{m}$

$\bf \: {x}^{0} = 1$

$\bf \: {x}^{m} = {x}^{n} ⇛ \: m = n$