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Solve-for-x-3-2x-18x-please-i-need-workings-




Question Number 10317 by Tawakalitu ayo mi last updated on 03/Feb/17
Solve for x.  3^(2x)  = 18x  please i need workings.
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}. \\ $$$$\mathrm{3}^{\mathrm{2x}} \:=\:\mathrm{18x} \\ $$$$\mathrm{please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{workings}. \\ $$
Answered by mrW1 last updated on 03/Feb/17
(3^2 )^x =18x  9^x =18x  e^(xln 9) =18x  1=((18)/(−ln 9))(−xln 9)e^(−xln 9)   (−xln 9)e^(−xln 9) =−((ln 9)/(18))  −xln 9=W(−((ln 9)/(18))) with W=Lambert W function  x=−((W(−((ln 9)/(18))))/(ln 9))=−((W(−0.122068))/(2.197225))  = { ((−((−0.140479)/(2.197225))=0.063935)),((−((−3.295837)/(2.197225))=1.5)) :}
$$\left(\mathrm{3}^{\mathrm{2}} \right)^{{x}} =\mathrm{18}{x} \\ $$$$\mathrm{9}^{{x}} =\mathrm{18}{x} \\ $$$${e}^{{x}\mathrm{ln}\:\mathrm{9}} =\mathrm{18}{x} \\ $$$$\mathrm{1}=\frac{\mathrm{18}}{−\mathrm{ln}\:\mathrm{9}}\left(−{x}\mathrm{ln}\:\mathrm{9}\right){e}^{−{x}\mathrm{ln}\:\mathrm{9}} \\ $$$$\left(−{x}\mathrm{ln}\:\mathrm{9}\right){e}^{−{x}\mathrm{ln}\:\mathrm{9}} =−\frac{\mathrm{ln}\:\mathrm{9}}{\mathrm{18}} \\ $$$$−{x}\mathrm{ln}\:\mathrm{9}={W}\left(−\frac{\mathrm{ln}\:\mathrm{9}}{\mathrm{18}}\right)\:{with}\:{W}={Lambert}\:{W}\:{function} \\ $$$${x}=−\frac{{W}\left(−\frac{\mathrm{ln}\:\mathrm{9}}{\mathrm{18}}\right)}{\mathrm{ln}\:\mathrm{9}}=−\frac{{W}\left(−\mathrm{0}.\mathrm{122068}\right)}{\mathrm{2}.\mathrm{197225}} \\ $$$$=\begin{cases}{−\frac{−\mathrm{0}.\mathrm{140479}}{\mathrm{2}.\mathrm{197225}}=\mathrm{0}.\mathrm{063935}}\\{−\frac{−\mathrm{3}.\mathrm{295837}}{\mathrm{2}.\mathrm{197225}}=\mathrm{1}.\mathrm{5}}\end{cases} \\ $$
Commented by Tawakalitu ayo mi last updated on 04/Feb/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by arge last updated on 04/Feb/17
    log 3^(2x) =log18x  2xlog 3=log 18+log x    log x=y    0.95x=1.26+y  x=((1.26+y)/(0.95))
$$ \\ $$$$ \\ $$$${log}\:\mathrm{3}^{\mathrm{2}{x}} ={log}\mathrm{18}{x} \\ $$$$\mathrm{2}{xlog}\:\mathrm{3}={log}\:\mathrm{18}+{log}\:{x} \\ $$$$ \\ $$$${log}\:{x}={y} \\ $$$$ \\ $$$$\mathrm{0}.\mathrm{95}{x}=\mathrm{1}.\mathrm{26}+{y} \\ $$$${x}=\frac{\mathrm{1}.\mathrm{26}+{y}}{\mathrm{0}.\mathrm{95}} \\ $$

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