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Question Number 200109 by sonukgindia last updated on 14/Nov/23

Answered by mr W last updated on 14/Nov/23

2^x =5−x 2^(x−5) =((5−x)/(32)) e^((x−5)ln 2) =((5−x)/(32)) (5−x)e^((5−x)ln 2) =32 (5−x)ln 2 e^((5−x)ln 2) =32ln 2 (5−x)ln 2 =W(32ln 2) ⇒x=5−((W(32ln 2))/(ln 2))≈5−((2.276558)/(ln 2))≈1.715621

$$\mathrm{2}^{{x}} =\mathrm{5}−{x} \\ $$$$\mathrm{2}^{{x}−\mathrm{5}} =\frac{\mathrm{5}−{x}}{\mathrm{32}} \\ $$$${e}^{\left({x}−\mathrm{5}\right)\mathrm{ln}\:\mathrm{2}} =\frac{\mathrm{5}−{x}}{\mathrm{32}} \\ $$$$\left(\mathrm{5}−{x}\right){e}^{\left(\mathrm{5}−{x}\right)\mathrm{ln}\:\mathrm{2}} =\mathrm{32} \\ $$$$\left(\mathrm{5}−{x}\right)\mathrm{ln}\:\mathrm{2}\:{e}^{\left(\mathrm{5}−{x}\right)\mathrm{ln}\:\mathrm{2}} =\mathrm{32ln}\:\mathrm{2} \\ $$$$\left(\mathrm{5}−{x}\right)\mathrm{ln}\:\mathrm{2}\:={W}\left(\mathrm{32ln}\:\mathrm{2}\right) \\ $$$$\Rightarrow{x}=\mathrm{5}−\frac{{W}\left(\mathrm{32ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}}\approx\mathrm{5}−\frac{\mathrm{2}.\mathrm{276558}}{\mathrm{ln}\:\mathrm{2}}\approx\mathrm{1}.\mathrm{715621} \\ $$

Commented by sonukgindia last updated on 14/Nov/23

solve next tetratioon question

$${solve}\:{next}\:{tetratioon}\:{question} \\ $$

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