{_((2^(2x) )^(1/3) +(2^(4x) )^(1/3) +4^x =5) ^(((2^x ))^(1/3) −2^x =2) }⇒2^x −8^x =? pleas solve thes


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Question Number 129841 by Adel last updated on 20/Jan/21
${_((2^(2x) )^(1/3) +(2^(4x) )^(1/3) +4^x =5) ^(((2^x ))^(1/3) −2^x =2) }⇒2^x −8^x =? pleas solve thes$
${_{\sqrt[3]{2^{2 x}} + \sqrt[3]{2^{4 x}} + 4^{x} = 5}^{\sqrt[3]{2^{x}} - 2^{x} = 2}} \Rightarrow 2^{x} - 8^{x} = ?$ $pleas solve thes$
Commented by MJS_new last updated on 20/Jan/21

$2^{x / 3} - 2^{x} = 2$ $2^{2 x / 3} + 2^{4 x / 3} + 4^{x} = 5$ $has no solution at all if {it}^{'} s a system of equations$ $two independent equations :$ $(1)$ $2^{x / 3} - 2^{x} = 2$ $t = 2^{x / 3} \Leftrightarrow x = \frac{3 \ln t}{\ln 2} \land t > 0$ $t^{3} - t + 2 = 0$ $t \approx - 1.52 \Rightarrow no real solution for x$ $(2)$ $2^{2 x / 3} + 2^{4 x / 3} + 4^{x} = 5$ $t = 2^{x / 3} \Leftrightarrow x = \frac{3 \ln t}{\ln 2} \land t > 0$ $t^{6} + t^{4} + t^{2} - 5 = 0$ $u = t^{2} + \frac{1}{3} \Leftrightarrow t = \sqrt{u - \frac{1}{3}} \land u ⩾ \frac{1}{3}$ $u^{3} + \frac{2}{3} u - \frac{142}{27} = 0$ $u \approx 1.61149641$ $\Rightarrow t \approx 1.13055874$ $\Rightarrow x \approx . 531107869$ $2^{x} - 8^{x} \approx - 1.57239847$
Commented by Adel last updated on 20/Jan/21

$thanks$
	Answered by liberty last updated on 20/Jan/21
	$2^x = y → { (((y)^(1/3) −y=2)),((((y)^(1/3) )^2 +((y)^(1/3) )^4 +y^2 =5)) :} substitute (y)^(1/3) = y+2 (y+2)^2 +(y+2)^4 +y^2 =5 (y+2)^2 {1+(y+2)^2 }+y^2 −5=0 let (y+2)^2 =z ∧y=(√z)−2 ⇒z(z+1)+z+4−4(√z) −5=0 ⇒z^2 +2z−4(√z)−1 =0 let (√z) = u ⇒u^4 +2u^2 −4u−1=0 ⇒(u^2 +2u+1)(u^2 −2u−1)=0 { (((u+1)^2 =0 ←not possible for u∈R)),(((u−1)^2 =2→u=1+(√2))) :} ⇒(√z) =1+(√2) , z = 3+2(√2) we get y = (√z)−2 = (√2)−1 now 2^x −8^x = y(1−y)(1+y) = ((√2)−1)(2−(√2))((√2)) = (√2) ((√2)−1)((√2)−1)(√2) = 2((√2)−1)^2 = 2(3−2(√(2 )))=6−4(√(2 ))$
	$2^{x} = y \to {\begin{cases} \sqrt[3]{y} - y = 2 \\ {(\sqrt[3]{y})}^{2} + {(\sqrt[3]{y})}^{4} + y^{2} = 5 \end{cases}$ $substitute \sqrt[3]{y} = y + 2$ ${(y + 2)}^{2} + {(y + 2)}^{4} + y^{2} = 5$ ${(y + 2)}^{2} {1 + {(y + 2)}^{2}} + y^{2} - 5 = 0$ $let {(y + 2)}^{2} = z \land y = \sqrt{z} - 2$ $\Rightarrow z (z + 1) + z + 4 - 4 \sqrt{z} - 5 = 0$ $\Rightarrow z^{2} + 2 z - 4 \sqrt{z} - 1 = 0$ $let \sqrt{z} = u$ $\Rightarrow u^{4} + {2 u}^{2} - 4 u - 1 = 0$ $\Rightarrow (u^{2} + 2 u + 1) (u^{2} - 2 u - 1) = 0$ ${\begin{cases} {(u + 1)}^{2} = 0 \leftarrow not possible for u \in R \\ {(u - 1)}^{2} = 2 \to u = 1 + \sqrt{2} \end{cases}$ $\Rightarrow \sqrt{z} = 1 + \sqrt{2}, z = 3 + 2 \sqrt{2}$ $we get y = \sqrt{z} - 2 = \sqrt{2} - 1$ $now 2^{x} - 8^{x} = y (1 - y) (1 + y)$ $= (\sqrt{2} - 1) (2 - \sqrt{2}) (\sqrt{2})$ $= \sqrt{2} (\sqrt{2} - 1) (\sqrt{2} - 1) \sqrt{2}$ $= 2 {(\sqrt{2} - 1)}^{2}$ $= 2 (3 - 2 \sqrt{2}) = 6 - 4 \sqrt{2}$
	Commented by Adel last updated on 20/Jan/21

	Commented by Adel last updated on 20/Jan/21

	$2^{x} - 8^{x} = ?$
	Commented by Adel last updated on 20/Jan/21

	$thanks$
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