if 4^x +4^(−x) =7 then 8^x +8^(−x) =?


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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
	$Another way 4^x +4^(−x) =7 ; 8^x +8^(−x) =? 2^(2x) +2^(−2x) =7 ⇒ { ((2^x (2^(2x) +2^(−2x) )=7(2^x ))),((2^(−x) (2^(2x) +2^(−2x) )=7(2^(−x) ))) :} ⇒ { ((2^(3x) +2^(−x) =7(2^x )....(i))),((2^x +2^(−3x) =7(2^(−x) )....(ii))) :} (i)+(ii): 8^x +8^(−x) +2^x +2^(−x) =7(2^x +2^(−x) ) 8^x +8^(−x) =6(2^x +2^(−x) ) 8^x +8^(−x) =6(3)=18 [∵ 2^x +2^(−x) =3]^★ ^★ 2^(2x) +2^(−2x) =7⇒(2^x +2^(−x) )^2 −2=7 ⇒(2^x +2^(−x) )^2 =3^2 ⇒2^x +2^(−x) =3$
	$A n o t h e r w a y$ $4^{x} + 4^{- x} = 7; 8^{x} + 8^{- x} = ?$ $2^{2 x} + 2^{- 2 x} = 7$ $\Rightarrow {\begin{cases} 2^{x} (2^{2 x} + 2^{- 2 x}) = 7 (2^{x}) \\ 2^{- x} (2^{2 x} + 2^{- 2 x}) = 7 (2^{- x}) \end{cases}$ $\Rightarrow {\begin{cases} 2^{3 x} + 2^{- x} = 7 (2^{x}) . . . . (i) \\ 2^{x} + 2^{- 3 x} = 7 (2^{- x}) . . . . (i i) \end{cases}$ $(i) + (i i) :$ $8^{x} + 8^{- x} + 2^{x} + 2^{- x} = 7 (2^{x} + 2^{- x})$ $8^{x} + 8^{- x} = 6 (2^{x} + 2^{- x})$ $8^{x} + 8^{- x} = 6 (3) = 18 {[∵ 2^{x} + 2^{- x} = 3]}^{★}$ $^{★} 2^{2 x} + 2^{- 2 x} = 7 \Rightarrow {(2^{x} + 2^{- x})}^{2} - 2 = 7$ $\Rightarrow {(2^{x} + 2^{- x})}^{2} = 3^{2} \Rightarrow 2^{x} + 2^{- x} = 3$
	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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