Question

2^x/1+2^x=1/4find the value of 8^x/1+8^x

Answer 1

Answer:

Thank you guys, for own maths problem share

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Answer 2

-47/64

Step-by-step explanation:

Correct Question

$\dfrac{1}{ {2}^{x} } + {2}^{x} = \dfrac{1}{4} \: \: \: ---(1)$

To find:

$\dfrac{1}{ {8}^{x} } + {8}^{x}$

Solution:

$\dfrac{1}{ {2}^{x} } + {2}^{x} = \dfrac{1}{4}$

[Cubing both side]

$({\dfrac{1}{ {2}^{x} } + {2}^{x} )}^{3} = {(\frac{1}{4})}^{3} \\ \\\dfrac{1}{ {2}^{3x} } + {2}^{3x} + 3. \frac{1}{ {2}^{x} }. {2}^{x}(\dfrac{1}{ {2}^{x} } + {2}^{x}) = \frac{1}{64} \\ \\ \frac{1}{ {8}^{x} } + {8}^{x} + 3. \frac{1}{ \cancel{{2}^{x}} }. \cancel{{2}^{x}}( \frac{1}{4}) = \frac{1}{64} \:\:\:\:[\text{from 1}] \\ \\ \frac{1}{ {8}^{x} } + {8}^{x} + \frac{3}{4} = \frac{1}{64} \\ \\ \frac{1}{ {8}^{x} } + {8}^{x} = \frac{1}{64} - \frac{3}{4} \\ \\ \frac{1}{ {8}^{x} } + {8}^{x} = \frac{1 - 48}{64} \\ \\ \frac{1}{ {8}^{x} } + {8}^{x} = - \frac{47}{64}$

Therefore the required answer is $- \dfrac{47}{64}$

${{\boxed{ \text{\blue{Important Algebra Formula}}}}}$

${(x + y)}^{2}={x}^{2}+{y}^{2}+2xy\\ \\{(x - y)}^{2}={x}^{2}+{y}^{2}-2xy \\\\{(x + y)}^{3}={x}^{3}+{y}^{3}+ 3xy(x + y) \\ \\(x - y)^{3}={x}^{3}-{y}^{3}- 3xy(x - y)$