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Question Number 40083 by tawa tawa last updated on 15/Jul/18

solve for x: 5^x + 5x = 140

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\:\:\mathrm{5}^{\mathrm{x}} \:+\:\mathrm{5x}\:=\:\mathrm{140} \\ $$

Commented by MrW3 last updated on 16/Jul/18

5^x =(28−x)×5 5^(x−28) =(28−x)×5^(−27) e^((x−28)ln 5) =(28−x)×5^(−27) 5^(27) ×ln 5=(28−x)ln 5 e^((28−x)ln 5) ⇒(28−x)ln 5=W(5^(27) ln 5) ⇒x=28−((W(5^(27) ln 5))/(ln 5))=28−((40.2359)/(ln 5))=3

$$\mathrm{5}^{{x}} =\left(\mathrm{28}−{x}\right)×\mathrm{5} \\ $$$$\mathrm{5}^{{x}−\mathrm{28}} =\left(\mathrm{28}−{x}\right)×\mathrm{5}^{−\mathrm{27}} \\ $$$${e}^{\left({x}−\mathrm{28}\right)\mathrm{ln}\:\mathrm{5}} =\left(\mathrm{28}−{x}\right)×\mathrm{5}^{−\mathrm{27}} \\ $$$$\mathrm{5}^{\mathrm{27}} ×\mathrm{ln}\:\mathrm{5}=\left(\mathrm{28}−{x}\right)\mathrm{ln}\:\mathrm{5}\:{e}^{\left(\mathrm{28}−{x}\right)\mathrm{ln}\:\mathrm{5}} \\ $$$$\Rightarrow\left(\mathrm{28}−{x}\right)\mathrm{ln}\:\mathrm{5}=\mathbb{W}\left(\mathrm{5}^{\mathrm{27}} \mathrm{ln}\:\mathrm{5}\right) \\ $$$$\Rightarrow{x}=\mathrm{28}−\frac{\mathbb{W}\left(\mathrm{5}^{\mathrm{27}} \mathrm{ln}\:\mathrm{5}\right)}{\mathrm{ln}\:\mathrm{5}}=\mathrm{28}−\frac{\mathrm{40}.\mathrm{2359}}{\mathrm{ln}\:\mathrm{5}}=\mathrm{3} \\ $$

Commented by tawa tawa last updated on 16/Jul/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by MJS last updated on 15/Jul/18

5^3 +5×3=140

$$\mathrm{5}^{\mathrm{3}} +\mathrm{5}×\mathrm{3}=\mathrm{140} \\ $$

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