Algebra Questions

Question Number 147945 by Gbenga last updated on 24/Jul/21

$$\mathrm{25}^{{x}} +\mathrm{5}{x}={e}^{\mathrm{5}} \:\left({can}\:{this}\:{be}\:{solved}\right) \\$$ $$\\$$

Commented bymr W last updated on 24/Jul/21

$${x}=\frac{{e}^{\mathrm{5}} }{\mathrm{5}}−\frac{\mathbb{W}\left(\mathrm{2}×\mathrm{5}^{\frac{\mathrm{2}{e}^{\mathrm{5}} }{\mathrm{5}}−\mathrm{1}} \mathrm{ln}\:\mathrm{5}\right)}{\mathrm{2ln}\:\mathrm{5}}\approx\mathrm{1}.\mathrm{536821} \\$$

Commented byGbenga last updated on 24/Jul/21

$${thanks} \\$$

Answered by Canebulok last updated on 24/Jul/21

$$\boldsymbol{{Solution}}: \\$$ $$\Rightarrow\:\mathrm{25}^{{x}} \:=\:{e}^{\mathrm{5}} −\mathrm{5}{x} \\$$ $$\Rightarrow\:\left({e}^{\mathrm{5}} −\mathrm{5}{x}\right)\centerdot\mathrm{5}^{−\mathrm{2}{x}} \:=\:\mathrm{1} \\$$ $$\Rightarrow\:\left({e}^{\mathrm{5}} −\mathrm{5}{x}\right){e}^{−\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)} \:=\:\mathrm{1} \\$$ $${By}\:{multiplying}\:{both}\:{sides}\:{by}\:\:\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}\:'', \\$$ $$\Rightarrow\:\left(\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}−\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)\right){e}^{−\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)} \:=\:\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}} \\$$ $$\Rightarrow\:\left(\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}−\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)\right){e}^{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}−\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)} \:=\:\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}\centerdot{e}^{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}} \\$$ $${By}\:{Lambert}\:{W}.{function}, \\$$ $$\Rightarrow\:\left(\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}−\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)\right)\:=\:{W}\left(\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}\centerdot{e}^{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}} \right) \\$$ $$\Rightarrow\:\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)\:=\:\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}−{W}\left(\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}\centerdot{e}^{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}} \right) \\$$ $$\Rightarrow\:{x}\:=\:\frac{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}−{W}\left(\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}\centerdot{e}^{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}} \right)}{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)} \\$$ $$\Rightarrow\:{x}\:=\:\frac{{e}^{\mathrm{5}} }{\mathrm{5}}−\frac{{W}\left(\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}\centerdot{e}^{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}} \right)}{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}\:\approx\:\mathrm{1}.\mathrm{536821146} \\$$ $$\: \\$$ $${Solution}\:{by}:\:{Kevin} \\$$

Commented byGbenga last updated on 24/Jul/21

$${nice}\:{solution} \\$$ $$\\$$