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Question Number 201152 by mathlove last updated on 01/Dec/23
if 4^x +4^(−x) =7 then 8^x +8^(−x) =?
$${if}\:\:\mathrm{4}^{{x}} +\mathrm{4}^{−{x}} =\mathrm{7} \\ $$$${then}\:\:\:\mathrm{8}^{{x}} +\mathrm{8}^{−{x}} =? \\ $$
Answered by BaliramKumar last updated on 01/Dec/23
(2^x )^2 + (1/((2^x )^2 )) +2= 7 +2 (2^x + (1/2^x ))^2 = 3^2 2^x + (1/2^x ) = 3 (2^x + (1/2^x ))^3 = 3^3 (2^x )^3 + (1/((2^x )^3 )) +3∙(2^x )∙(1/((2^x )))(2^x + (1/2^x ))= 27 2^(3x) +(1/2^(3x) )+3(3) = 27 2^(3x) +(1/2^(3x) ) = 27 − 9 8^x + 8^(−x) = 18
$$\left(\mathrm{2}^{\mathrm{x}} \right)^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\left(\mathrm{2}^{\mathrm{x}} \right)^{\mathrm{2}} }\:+\mathrm{2}=\:\mathrm{7}\:+\mathrm{2} \\ $$$$\left(\mathrm{2}^{\mathrm{x}} \:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{x}} }\right)^{\mathrm{2}} \:=\:\mathrm{3}^{\mathrm{2}} \\ $$$$\mathrm{2}^{\mathrm{x}} \:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{x}} }\:=\:\mathrm{3} \\ $$$$\left(\mathrm{2}^{\mathrm{x}} \:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{x}} }\right)^{\mathrm{3}} \:=\:\mathrm{3}^{\mathrm{3}} \\ $$$$\left(\mathrm{2}^{\mathrm{x}} \right)^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\left(\mathrm{2}^{\mathrm{x}} \right)^{\mathrm{3}} }\:+\mathrm{3}\centerdot\cancel{\left(\mathrm{2}^{\mathrm{x}} \right)}\centerdot\frac{\mathrm{1}}{\cancel{\left(\mathrm{2}^{\mathrm{x}} \right)}}\left(\mathrm{2}^{\mathrm{x}} \:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{x}} }\right)=\:\mathrm{27} \\ $$$$\mathrm{2}^{\mathrm{3x}} +\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3x}} }+\mathrm{3}\left(\mathrm{3}\right)\:=\:\mathrm{27} \\ $$$$\mathrm{2}^{\mathrm{3x}} +\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3x}} }\:=\:\mathrm{27}\:−\:\mathrm{9}\: \\ $$$$\mathrm{8}^{\mathrm{x}} \:+\:\mathrm{8}^{−\mathrm{x}} \:=\:\mathrm{18} \\ $$
Commented by mathlove last updated on 01/Dec/23
thanks
$${thanks} \\ $$
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
$$\mathcal{A}{nother}\:{way} \\ $$$$\mathrm{4}^{{x}} +\mathrm{4}^{−{x}} =\mathrm{7}\:;\:\:\:\mathrm{8}^{{x}} +\mathrm{8}^{−{x}} =? \\ $$$$\mathrm{2}^{\mathrm{2}{x}} +\mathrm{2}^{−\mathrm{2}{x}} =\mathrm{7} \\ $$$$\Rightarrow\begin{cases}{\mathrm{2}^{{x}} \left(\mathrm{2}^{\mathrm{2}{x}} +\mathrm{2}^{−\mathrm{2}{x}} \right)=\mathrm{7}\left(\mathrm{2}^{{x}} \right)}\\{\mathrm{2}^{−{x}} \left(\mathrm{2}^{\mathrm{2}{x}} +\mathrm{2}^{−\mathrm{2}{x}} \right)=\mathrm{7}\left(\mathrm{2}^{−{x}} \right)}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{2}^{\mathrm{3}{x}} +\mathrm{2}^{−{x}} =\mathrm{7}\left(\mathrm{2}^{{x}} \right)….\left({i}\right)}\\{\mathrm{2}^{{x}} +\mathrm{2}^{−\mathrm{3}{x}} =\mathrm{7}\left(\mathrm{2}^{−{x}} \right)….\left({ii}\right)}\end{cases} \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\:\:\:\mathrm{8}^{{x}} +\mathrm{8}^{−{x}} +\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} =\mathrm{7}\left(\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} \right) \\ $$$$\:\:\:\mathrm{8}^{{x}} +\mathrm{8}^{−{x}} =\mathrm{6}\left(\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} \right) \\ $$$$\:\:\:\mathrm{8}^{{x}} +\mathrm{8}^{−{x}} =\mathrm{6}\left(\mathrm{3}\right)=\mathrm{18}\:\:\left[\because\:\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} =\mathrm{3}\right]^{\bigstar} \\ $$$$\: \\ $$$$\:^{\bigstar} \:\mathrm{2}^{\mathrm{2}{x}} +\mathrm{2}^{−\mathrm{2}{x}} =\mathrm{7}\Rightarrow\left(\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} \right)^{\mathrm{2}} −\mathrm{2}=\mathrm{7} \\ $$$$\Rightarrow\left(\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} \right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} \Rightarrow\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} =\mathrm{3} \\ $$
Commented by mathlove last updated on 01/Dec/23
thanks
$${thanks} \\ $$
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^(−2x) +2^(−x) 2^(2x) +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✓
$$\mathrm{4}^{{x}} +\mathrm{4}^{−{x}} =\mathrm{7}\:;\:\mathrm{8}^{{x}} +\mathrm{8}^{−{x}} =? \\ $$$$\mathrm{2}^{{x}} \left(\mathrm{4}^{{x}} +\mathrm{4}^{−{x}} \right)+\mathrm{2}^{−{x}} \left(\mathrm{4}^{{x}} +\mathrm{4}^{−{x}} \right)=\mathrm{2}^{{x}} \left(\mathrm{7}\right)+\mathrm{2}^{−{x}} \left(\mathrm{7}\right) \\ $$$$\mathrm{8}^{{x}} +\mathrm{2}^{{x}} \mathrm{2}^{−\mathrm{2}{x}} +\mathrm{2}^{−{x}} \mathrm{2}^{\mathrm{2}{x}} +\mathrm{8}^{−{x}} =\mathrm{7}\left(\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} \right) \\ $$$$\mathrm{8}^{{x}} +\mathrm{8}^{−{x}} +\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} =\mathrm{7}\left(\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} \right) \\ $$$$\mathrm{8}^{{x}} +\mathrm{8}^{−{x}} =\mathrm{6}\left(\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} \right)=\mathrm{6}\left(\mathrm{3}\right)=\mathrm{18}\checkmark \\ $$
Answered by esmaeil last updated on 02/Dec/23
4^x +4^(−x) =(2^x +2^(−x) )^2 −2=7→ 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
$$\mathrm{4}^{{x}} +\mathrm{4}^{−{x}} =\left(\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} \right)^{\mathrm{2}} −\mathrm{2}=\mathrm{7}\rightarrow \\ $$$$\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} =\mathrm{3} \\ $$$$\mathrm{8}^{{x}} +\mathrm{8}^{−{x}} =\left(\mathrm{2}^{{x}} \right)^{\mathrm{3}} +\left(\mathrm{2}^{−{x}} \right)^{\mathrm{3}} = \\ $$$$\left(\mathrm{2}^{{x}} +\mathrm{2}^{−{x}} \right)\left(\mathrm{4}^{{x}} +\mathrm{4}^{−{x}} −\mathrm{1}\right)=\mathrm{18} \\ $$

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