Question Number 187336 by Humble last updated on 16/Feb/23
$${solve}\:{for}\:{x}\:{if} \\ $$$${X}^{{x}} \bullet\mathrm{5}^{{x}} −\mathrm{5}^{\mathrm{2}+{x}} =\mathrm{0} \\ $$$$ \\ $$
Answered by mr W last updated on 16/Feb/23
$${x}^{{x}} ×\mathrm{5}^{{x}} −\mathrm{25}×\mathrm{5}^{{x}} =\mathrm{0} \\ $$$$\left({x}^{{x}} −\mathrm{25}\right)\mathrm{5}^{{x}} =\mathrm{0} \\ $$$$\mathrm{5}^{{x}} \neq\mathrm{0} \\ $$$$\Rightarrow{x}^{{x}} −\mathrm{25}=\mathrm{0} \\ $$$$\Rightarrow{x}^{{x}} =\mathrm{25} \\ $$$$\Rightarrow{x}\mathrm{ln}\:{x}=\mathrm{ln}\:\mathrm{25} \\ $$$$\Rightarrow\left(\mathrm{ln}\:{x}\right){e}^{\mathrm{ln}\:{x}} =\mathrm{ln}\:\mathrm{25} \\ $$$$\Rightarrow\mathrm{ln}\:{x}={W}\left(\mathrm{ln}\:\mathrm{25}\right) \\ $$$$\Rightarrow{x}={e}^{{W}\left(\mathrm{ln}\:\mathrm{25}\right)} =\frac{\mathrm{ln}\:\mathrm{25}}{{W}\left(\mathrm{ln}\:\mathrm{25}\right)} \\ $$$$\:\:\:\:\:\:\:\:\approx\frac{\mathrm{ln}\:\mathrm{25}}{\mathrm{1}.\mathrm{086276}}\approx\mathrm{2}.\mathrm{96322}\:\checkmark\: \\ $$
Commented by Humble last updated on 16/Feb/23
$${Excellent}\:{solution}!\:{thank}\:{you},\:{sir} \\ $$