Question Number 200109 by sonukgindia last updated on 14/Nov/23
Answered by mr W last updated on 14/Nov/23
$$\mathrm{2}^{{x}} =\mathrm{5}−{x} \\ $$$$\mathrm{2}^{{x}−\mathrm{5}} =\frac{\mathrm{5}−{x}}{\mathrm{32}} \\ $$$${e}^{\left({x}−\mathrm{5}\right)\mathrm{ln}\:\mathrm{2}} =\frac{\mathrm{5}−{x}}{\mathrm{32}} \\ $$$$\left(\mathrm{5}−{x}\right){e}^{\left(\mathrm{5}−{x}\right)\mathrm{ln}\:\mathrm{2}} =\mathrm{32} \\ $$$$\left(\mathrm{5}−{x}\right)\mathrm{ln}\:\mathrm{2}\:{e}^{\left(\mathrm{5}−{x}\right)\mathrm{ln}\:\mathrm{2}} =\mathrm{32ln}\:\mathrm{2} \\ $$$$\left(\mathrm{5}−{x}\right)\mathrm{ln}\:\mathrm{2}\:={W}\left(\mathrm{32ln}\:\mathrm{2}\right) \\ $$$$\Rightarrow{x}=\mathrm{5}−\frac{{W}\left(\mathrm{32ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}}\approx\mathrm{5}−\frac{\mathrm{2}.\mathrm{276558}}{\mathrm{ln}\:\mathrm{2}}\approx\mathrm{1}.\mathrm{715621} \\ $$
Commented by sonukgindia last updated on 14/Nov/23
$${solve}\:{next}\:{tetratioon}\:{question} \\ $$