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Question Number 189576 by normans last updated on 18/Mar/23
2^x + 2^x −4 + x = −(1/2) x = ???
$$ \\ $$$$\:\:\:\:\:\mathrm{2}^{\boldsymbol{{x}}} \:+\:\mathrm{2}^{\boldsymbol{{x}}} −\mathrm{4}\:+\:\boldsymbol{{x}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\boldsymbol{{x}}\:=\:??? \\ $$$$ \\ $$
Answered by Gbenga last updated on 18/Mar/23
solution 2^x +2^x −4+x=−(1/2) 2(2^x )+x=−(1/2)+4=(7/2) 2^(x+1) +x=(7/2) 2^(x+1) =(7/2)−x (2^(x+1) )2^(−x) =((7/2)−x)2^(−x) 2=((7/2)−x)2^(−x) 2(2^(7/2) )=((7/2)−x)2^((7/2)−x) 16(√2)=((7/2)−x)e^(((7/2)−x)ln(2)) 16(√2)ln(2)=((7/2)−x)ln(2)e^(((7/2)−x)ln(2)) W(16(√2)ln(2))=W(((7/2)−x)ln(2)e^(((7/2)−x)ln(2)) ) ((7/2)−x)ln(2)=W(16(√2)ln(2)) ((7/2)−x)=((W(16(√2)ln(2)))/(ln(2))) x=(7/2)−((W(16(√2)ln(2)))/(ln(2)))≈0.5571929518106867... ★Small Laplace★
$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{solution}} \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} +\mathrm{2}^{\boldsymbol{\mathrm{x}}} −\mathrm{4}+\boldsymbol{\mathrm{x}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{2}^{\boldsymbol{\mathrm{x}}} \right)+\boldsymbol{\mathrm{x}}=−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{4}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}+\mathrm{1}} +\boldsymbol{\mathrm{x}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}+\mathrm{1}} =\frac{\mathrm{7}}{\mathrm{2}}−\boldsymbol{\mathrm{x}} \\ $$$$\left(\mathrm{2}^{\boldsymbol{\mathrm{x}}+\mathrm{1}} \right)\mathrm{2}^{−\boldsymbol{\mathrm{x}}} =\left(\frac{\mathrm{7}}{\mathrm{2}}−\boldsymbol{\mathrm{x}}\right)\mathrm{2}^{−\boldsymbol{\mathrm{x}}} \\ $$$$\mathrm{2}=\left(\frac{\mathrm{7}}{\mathrm{2}}−\boldsymbol{\mathrm{x}}\right)\mathrm{2}^{−\boldsymbol{\mathrm{x}}} \\ $$$$\mathrm{2}\left(\mathrm{2}^{\frac{\mathrm{7}}{\mathrm{2}}} \right)=\left(\frac{\mathrm{7}}{\mathrm{2}}−\boldsymbol{\mathrm{x}}\right)\mathrm{2}^{\frac{\mathrm{7}}{\mathrm{2}}−\boldsymbol{\mathrm{x}}} \\ $$$$\mathrm{16}\sqrt{\mathrm{2}}=\left(\frac{\mathrm{7}}{\mathrm{2}}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{e}}^{\left(\frac{\mathrm{7}}{\mathrm{2}}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)} \\ $$$$\mathrm{16}\sqrt{\mathrm{2}}\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)=\left(\frac{\mathrm{7}}{\mathrm{2}}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)\boldsymbol{\mathrm{e}}^{\left(\frac{\mathrm{7}}{\mathrm{2}}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)} \\ $$$$\boldsymbol{\mathrm{W}}\left(\mathrm{16}\sqrt{\mathrm{2}}\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)\right)=\boldsymbol{\mathrm{W}}\left(\left(\frac{\mathrm{7}}{\mathrm{2}}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)\boldsymbol{\mathrm{e}}^{\left(\frac{\mathrm{7}}{\mathrm{2}}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)} \right) \\ $$$$\left(\frac{\mathrm{7}}{\mathrm{2}}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)=\boldsymbol{\mathrm{W}}\left(\mathrm{16}\sqrt{\mathrm{2}}\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)\right) \\ $$$$\left(\frac{\mathrm{7}}{\mathrm{2}}−\boldsymbol{\mathrm{x}}\right)=\frac{\boldsymbol{\mathrm{W}}\left(\mathrm{16}\sqrt{\mathrm{2}}\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)\right)}{\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)} \\ $$$$\boldsymbol{\mathrm{x}}=\frac{\mathrm{7}}{\mathrm{2}}−\frac{\boldsymbol{\mathrm{W}}\left(\mathrm{16}\sqrt{\mathrm{2}}\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)\right)}{\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)}\approx\mathrm{0}.\mathrm{5571929518106867}… \\ $$$$\bigstar\boldsymbol{{Small}}\:\boldsymbol{{Laplace}}\bigstar \\ $$

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