Math, asked by csruchiaggarwal, 10 months ago

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

Answers

Answered by santlalsharma20195
0

Answer:

Thank you guys, for own maths problem share

$$$$$$$$$$========#######:@@@@@@

Attachments:
Answered by tahseen619
4

-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)

Similar questions