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25-x-5x-e-5-can-this-be-solved-

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Question Number 147945 by Gbenga last updated on 24/Jul/21
25^x +5x=e^5 (can this be solved)
$$\mathrm{25}^{{x}} +\mathrm{5}{x}={e}^{\mathrm{5}} \:\left({can}\:{this}\:{be}\:{solved}\right) \\ $$$$ \\ $$
Commented by mr W last updated on 24/Jul/21
x=(e^5 /5)−((W(2×5^(((2e^5 )/5)−1) ln 5))/(2ln 5))≈1.536821
$${x}=\frac{{e}^{\mathrm{5}} }{\mathrm{5}}−\frac{\mathbb{W}\left(\mathrm{2}×\mathrm{5}^{\frac{\mathrm{2}{e}^{\mathrm{5}} }{\mathrm{5}}−\mathrm{1}} \mathrm{ln}\:\mathrm{5}\right)}{\mathrm{2ln}\:\mathrm{5}}\approx\mathrm{1}.\mathrm{536821} \\ $$
Commented by Gbenga last updated on 24/Jul/21
thanks
$${thanks} \\ $$
Answered by Canebulok last updated on 24/Jul/21
Solution: ⇒ 25^x = e^5 −5x ⇒ (e^5 −5x)∙5^(−2x) = 1 ⇒ (e^5 −5x)e^(−2x∙ln(5)) = 1 By multiplying both sides by “ ((2∙ln(5))/5) ”, ⇒ (((2e^5 ∙ln(5))/5)−2x∙ln(5))e^(−2x∙ln(5)) = ((2∙ln(5))/5) ⇒ (((2e^5 ∙ln(5))/5)−2x∙ln(5))e^(((2e^5 ∙ln(5))/5)−2x∙ln(5)) = ((2∙ln(5))/5)∙e^((2e^5 ∙ln(5))/5) By Lambert W.function, ⇒ (((2e^5 ∙ln(5))/5)−2x∙ln(5)) = W(((2∙ln(5))/5)∙e^((2e^5 ∙ln(5))/5) ) ⇒ 2x∙ln(5) = ((2e^5 ∙ln(5))/5)−W(((2∙ln(5))/5)∙e^((2e^5 ∙ln(5))/5) ) ⇒ x = ((((2e^5 ∙ln(5))/5)−W(((2∙ln(5))/5)∙e^((2e^5 ∙ln(5))/5) ))/(2∙ln(5))) ⇒ x = (e^5 /5)−((W(((2∙ln(5))/5)∙e^((2e^5 ∙ln(5))/5) ))/(2∙ln(5))) ≈ 1.536821146 Solution by: Kevin
$$\boldsymbol{{Solution}}: \\ $$$$\Rightarrow\:\mathrm{25}^{{x}} \:=\:{e}^{\mathrm{5}} −\mathrm{5}{x} \\ $$$$\Rightarrow\:\left({e}^{\mathrm{5}} −\mathrm{5}{x}\right)\centerdot\mathrm{5}^{−\mathrm{2}{x}} \:=\:\mathrm{1} \\ $$$$\Rightarrow\:\left({e}^{\mathrm{5}} −\mathrm{5}{x}\right){e}^{−\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)} \:=\:\mathrm{1} \\ $$$${By}\:{multiplying}\:{both}\:{sides}\:{by}\:“\:\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}\:'', \\ $$$$\Rightarrow\:\left(\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}−\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)\right){e}^{−\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)} \:=\:\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}} \\ $$$$\Rightarrow\:\left(\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}−\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)\right){e}^{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}−\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)} \:=\:\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}\centerdot{e}^{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}} \\ $$$${By}\:{Lambert}\:{W}.{function}, \\ $$$$\Rightarrow\:\left(\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}−\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)\right)\:=\:{W}\left(\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}\centerdot{e}^{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}} \right) \\ $$$$\Rightarrow\:\mathrm{2}{x}\centerdot{ln}\left(\mathrm{5}\right)\:=\:\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}−{W}\left(\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}\centerdot{e}^{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}} \right) \\ $$$$\Rightarrow\:{x}\:=\:\frac{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}−{W}\left(\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}\centerdot{e}^{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}} \right)}{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)} \\ $$$$\Rightarrow\:{x}\:=\:\frac{{e}^{\mathrm{5}} }{\mathrm{5}}−\frac{{W}\left(\frac{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}\centerdot{e}^{\frac{\mathrm{2}{e}^{\mathrm{5}} \centerdot{ln}\left(\mathrm{5}\right)}{\mathrm{5}}} \right)}{\mathrm{2}\centerdot{ln}\left(\mathrm{5}\right)}\:\approx\:\mathrm{1}.\mathrm{536821146} \\ $$$$\: \\ $$$${Solution}\:{by}:\:{Kevin} \\ $$
Commented by Gbenga last updated on 24/Jul/21
nice solution
$${nice}\:{solution} \\ $$$$ \\ $$

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