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Question Number 194176 by Skabetix last updated on 29/Jun/23

Hello everyone I try to solve 4^(x+1) +2^(2−x) =65 Thx in advance

$${Hello}\:{everyone} \\ $$$${I}\:{try}\:{to}\:{solve}\:\mathrm{4}^{{x}+\mathrm{1}} +\mathrm{2}^{\mathrm{2}−{x}} =\mathrm{65} \\ $$$${Thx}\:{in}\:{advance} \\ $$

Answered by Skabetix last updated on 29/Jun/23

I found x=2 but there is an other solution

$${I}\:{found}\:{x}=\mathrm{2}\:{but}\:{there}\:{is}\:{an}\:{other}\:{solution} \\ $$

Answered by mr W last updated on 29/Jun/23

4(2^x )^2 +(4/2^x )=65 4(2^x )^3 −65(2^x )+4=0 let u=2^x 4u^3 −65u+4=0 4u^3 −16u^2 +16u^2 −64u−u+4=0 (u−4)(4u^2 +16u−1)=0 ⇒u=4=2^x ⇒x=2 ✓ ⇒u=((−4+(√(17)))/2)=2^x ⇒x=log_2 ((−4+(√(17)))/2) ✓

$$\mathrm{4}\left(\mathrm{2}^{{x}} \right)^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{2}^{{x}} }=\mathrm{65} \\ $$$$\mathrm{4}\left(\mathrm{2}^{{x}} \right)^{\mathrm{3}} −\mathrm{65}\left(\mathrm{2}^{{x}} \right)+\mathrm{4}=\mathrm{0} \\ $$$${let}\:{u}=\mathrm{2}^{{x}} \\ $$$$\mathrm{4}{u}^{\mathrm{3}} −\mathrm{65}{u}+\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{4}{u}^{\mathrm{3}} −\mathrm{16}{u}^{\mathrm{2}} +\mathrm{16}{u}^{\mathrm{2}} −\mathrm{64}{u}−{u}+\mathrm{4}=\mathrm{0} \\ $$$$\left({u}−\mathrm{4}\right)\left(\mathrm{4}{u}^{\mathrm{2}} +\mathrm{16}{u}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{4}=\mathrm{2}^{{x}} \:\Rightarrow{x}=\mathrm{2}\:\checkmark \\ $$$$\Rightarrow{u}=\frac{−\mathrm{4}+\sqrt{\mathrm{17}}}{\mathrm{2}}=\mathrm{2}^{{x}} \:\Rightarrow{x}=\mathrm{log}_{\mathrm{2}} \:\frac{−\mathrm{4}+\sqrt{\mathrm{17}}}{\mathrm{2}}\:\checkmark \\ $$

Answered by MM42 last updated on 29/Jun/23

2^x (2^(2x+2) +2^(2−x) =65) 4×2^(3x) +4=65×2^x 2^x =y 4y^3 −65y+4=0 (y−4)(4y^2 +16y−1)=0 y=4=2^x ⇒x=2 y=−8±(√(68))=−8±2(√(17))=2(−4±(√(17))) 2^x =2(−4±(√(17)))⇒x=2+log_2 (−4±(√(17)))

$$\mathrm{2}^{{x}} \left(\mathrm{2}^{\mathrm{2}{x}+\mathrm{2}} +\mathrm{2}^{\mathrm{2}−{x}} =\mathrm{65}\right) \\ $$$$\mathrm{4}×\mathrm{2}^{\mathrm{3}{x}} +\mathrm{4}=\mathrm{65}×\mathrm{2}^{{x}} \\ $$$$\mathrm{2}^{{x}} ={y} \\ $$$$\mathrm{4}{y}^{\mathrm{3}} −\mathrm{65}{y}+\mathrm{4}=\mathrm{0} \\ $$$$\left({y}−\mathrm{4}\right)\left(\mathrm{4}{y}^{\mathrm{2}} +\mathrm{16}{y}−\mathrm{1}\right)=\mathrm{0} \\ $$$${y}=\mathrm{4}=\mathrm{2}^{{x}} \Rightarrow{x}=\mathrm{2} \\ $$$${y}=−\mathrm{8}\pm\sqrt{\mathrm{68}}=−\mathrm{8}\pm\mathrm{2}\sqrt{\mathrm{17}}=\mathrm{2}\left(−\mathrm{4}\pm\sqrt{\mathrm{17}}\right) \\ $$$$\mathrm{2}^{{x}} =\mathrm{2}\left(−\mathrm{4}\pm\sqrt{\mathrm{17}}\right)\Rightarrow{x}=\mathrm{2}+{log}_{\mathrm{2}} \left(−\mathrm{4}\pm\sqrt{\mathrm{17}}\right) \\ $$$$ \\ $$

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