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Question Number 13152 by tawa tawa last updated on 15/May/17

Solve: 5^(log(x)) + logx^5 = 25

$$\mathrm{Solve}:\:\:\mathrm{5}^{\mathrm{log}\left(\mathrm{x}\right)} \:+\:\mathrm{logx}^{\mathrm{5}} \:=\:\mathrm{25} \\ $$

Answered by mrW1 last updated on 16/May/17

log x=t 5^t +5t=25 5^t =5(5−t) 5^(t−5) =((5−t)/(625)) e^((t−5)ln 5) =((5−t)/(625)) (5−t)ln 5e^((t−5)ln 5) =625×ln 5 (5−t)ln 5=W(625×ln 5) t=5−((W(625×ln 5))/(ln 5))=log x ⇒x=10^(5−((W(625×ln 5))/(ln 5))) ≈10^(5−((5.254544)/(ln 5))) ≈54.346

$$\mathrm{log}\:{x}={t} \\ $$$$\mathrm{5}^{{t}} +\mathrm{5}{t}=\mathrm{25} \\ $$$$\mathrm{5}^{{t}} =\mathrm{5}\left(\mathrm{5}−{t}\right) \\ $$$$\mathrm{5}^{{t}−\mathrm{5}} =\frac{\mathrm{5}−{t}}{\mathrm{625}} \\ $$$${e}^{\left({t}−\mathrm{5}\right)\mathrm{ln}\:\mathrm{5}} =\frac{\mathrm{5}−{t}}{\mathrm{625}} \\ $$$$\left(\mathrm{5}−{t}\right)\mathrm{ln}\:\mathrm{5}{e}^{\left({t}−\mathrm{5}\right)\mathrm{ln}\:\mathrm{5}} =\mathrm{625}×\mathrm{ln}\:\mathrm{5} \\ $$$$\left(\mathrm{5}−{t}\right)\mathrm{ln}\:\mathrm{5}={W}\left(\mathrm{625}×\mathrm{ln}\:\mathrm{5}\right) \\ $$$${t}=\mathrm{5}−\frac{{W}\left(\mathrm{625}×\mathrm{ln}\:\mathrm{5}\right)}{\mathrm{ln}\:\mathrm{5}}=\mathrm{log}\:{x} \\ $$$$\Rightarrow{x}=\mathrm{10}^{\mathrm{5}−\frac{{W}\left(\mathrm{625}×\mathrm{ln}\:\mathrm{5}\right)}{\mathrm{ln}\:\mathrm{5}}} \approx\mathrm{10}^{\mathrm{5}−\frac{\mathrm{5}.\mathrm{254544}}{\mathrm{ln}\:\mathrm{5}}} \approx\mathrm{54}.\mathrm{346} \\ $$

Commented by tawa tawa last updated on 15/May/17

Wow, God bless you sir.

$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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