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Question Number 9875 by j.masanja06@gmail.com last updated on 11/Jan/17

solve the value of x. 5logx=(x/4)

$$\mathrm{solve}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$$$\mathrm{5logx}=\frac{\mathrm{x}}{\mathrm{4}} \\ $$

Answered by mrW1 last updated on 12/Jan/17

5log x=(x/4) log x=(x/(20)) ((ln x)/(ln 10))=(x/(20)) ln x=((ln 10)/(20))x with a=−((ln 10)/(20))≈−0.115129 ln x=−ax x=e^(−ax) =(1/e^(ax) ) xe^(ax) =1 (ax)e^(ax) =a ⇒ax=W(a) with W=Lambert W function x=((W(a))/a)=((W(−((ln 10)/(20))))/(−((ln 10)/(20))))≈((W(−0.115129))/(−0.115129)) ≈((−0.13128)/(−0.115129))≈1.140286 or ≈((−3.3794)/(−0.115128))≈29.35316

$$\mathrm{5log}\:{x}=\frac{{x}}{\mathrm{4}} \\ $$$$\mathrm{log}\:{x}=\frac{{x}}{\mathrm{20}} \\ $$$$\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{10}}=\frac{{x}}{\mathrm{20}} \\ $$$$\mathrm{ln}\:{x}=\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{20}}{x} \\ $$$${with}\:{a}=−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{20}}\approx−\mathrm{0}.\mathrm{115129} \\ $$$$\mathrm{ln}\:{x}=−{ax} \\ $$$${x}={e}^{−{ax}} =\frac{\mathrm{1}}{{e}^{{ax}} } \\ $$$${xe}^{{ax}} =\mathrm{1} \\ $$$$\left({ax}\right){e}^{{ax}} ={a} \\ $$$$\Rightarrow{ax}={W}\left({a}\right) \\ $$$${with}\:{W}={Lambert}\:{W}\:{function} \\ $$$${x}=\frac{{W}\left({a}\right)}{{a}}=\frac{{W}\left(−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{20}}\right)}{−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{20}}}\approx\frac{{W}\left(−\mathrm{0}.\mathrm{115129}\right)}{−\mathrm{0}.\mathrm{115129}} \\ $$$$\approx\frac{−\mathrm{0}.\mathrm{13128}}{−\mathrm{0}.\mathrm{115129}}\approx\mathrm{1}.\mathrm{140286} \\ $$$${or} \\ $$$$\approx\frac{−\mathrm{3}.\mathrm{3794}}{−\mathrm{0}.\mathrm{115128}}\approx\mathrm{29}.\mathrm{35316} \\ $$

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