if-4-x-4-x-7-then-8-x-8-x-

Question Number 201152 by mathlove last updated on 01/Dec/23

i f 4^{x} + 4^{- x} = 7

t h e n 8^{x} + 8^{- x} = ?

Answered by BaliramKumar last updated on 01/Dec/23

{(2^{x})}^{2} + \frac{1}{{(2^{x})}^{2}} + 2 = 7 + 2

{(2^{x} + \frac{1}{2^{x}})}^{2} = 3^{2}

2^{x} + \frac{1}{2^{x}} = 3

{(2^{x} + \frac{1}{2^{x}})}^{3} = 3^{3}

{(2^{x})}^{3} + \frac{1}{{(2^{x})}^{3}} + 3 \cdot (2^{x}) \cdot \frac{1}{(2^{x})} (2^{x} + \frac{1}{2^{x}}) = 27

2^{3 x} + \frac{1}{2^{3 x}} + 3 (3) = 27

2^{3 x} + \frac{1}{2^{3 x}} = 27 - 9

8^{x} + 8^{- x} = 18

Commented by mathlove last updated on 01/Dec/23

t h a n k s

Answered by Rasheed.Sindhi last updated on 01/Dec/23

A n o t h e r w a y

4^{x} + 4^{- x} = 7; 8^{x} + 8^{- x} = ?

2^{2 x} + 2^{- 2 x} = 7

\Rightarrow {\begin{cases} 2^{x} (2^{2 x} + 2^{- 2 x}) = 7 (2^{x}) \\ 2^{- x} (2^{2 x} + 2^{- 2 x}) = 7 (2^{- x}) \end{cases}

\Rightarrow {\begin{cases} 2^{3 x} + 2^{- x} = 7 (2^{x}) \dots . (i) \\ 2^{x} + 2^{- 3 x} = 7 (2^{- x}) \dots . (i i) \end{cases}

(i) + (i i) :

8^{x} + 8^{- x} + 2^{x} + 2^{- x} = 7 (2^{x} + 2^{- x})

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

8^{x} + 8^{- x} = 6 (3) = 18 {[∵ 2^{x} + 2^{- x} = 3]}^{★}

^{★} 2^{2 x} + 2^{- 2 x} = 7 \Rightarrow {(2^{x} + 2^{- x})}^{2} - 2 = 7

\Rightarrow {(2^{x} + 2^{- x})}^{2} = 3^{2} \Rightarrow 2^{x} + 2^{- x} = 3

Commented by mathlove last updated on 01/Dec/23

t h a n k s

Answered by Rasheed.Sindhi last updated on 01/Dec/23

4^{x} + 4^{- x} = 7; 8^{x} + 8^{- x} = ?

2^{x} (4^{x} + 4^{- x}) + 2^{- x} (4^{x} + 4^{- x}) = 2^{x} (7) + 2^{- x} (7)

8^{x} + 2^{x} 2^{- 2 x} + 2^{- x} 2^{2 x} + 8^{- x} = 7 (2^{x} + 2^{- x})

8^{x} + 8^{- x} + 2^{x} + 2^{- x} = 7 (2^{x} + 2^{- x})

8^{x} + 8^{- x} = 6 (2^{x} + 2^{- x}) = 6 (3) = 18 ✓

Answered by esmaeil last updated on 02/Dec/23

4^{x} + 4^{- x} = {(2^{x} + 2^{- x})}^{2} - 2 = 7 \to

2^{x} + 2^{- x} = 3

8^{x} + 8^{- x} = {(2^{x})}^{3} + {(2^{- x})}^{3} =

(2^{x} + 2^{- x}) (4^{x} + 4^{- x} - 1) = 18

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