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Question Number 152571 by Gbenga last updated on 29/Aug/21

2log4^x +9logx^4 =9 find x

$$\:\mathrm{2}\boldsymbol{{log}}\mathrm{4}^{\boldsymbol{{x}}} +\mathrm{9}\boldsymbol{{logx}}^{\mathrm{4}} =\mathrm{9} \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{x}} \\ $$

Answered by mr W last updated on 29/Aug/21

log (4^(2x) ×x^(36) )=9 4^(2x) ×x^(36) =10^9 2^(x/9) ×x=±10^(1/4) e^((x/9)ln 2) ×x=±10^(1/4) e^((x/9)ln 2) ×((xln 2)/9)=±10^(1/4) ×((ln 2)/9) ((xln 2)/9)=W(±10^(1/4) ×((ln 2)/9)) ⇒x=(9/(ln 2))W(±10^(1/4) ×((ln 2)/9)) = { (((9/(ln 2))W(−10^(1/4) ×((ln 2)/9))= { (((9/(ln 2))×(−3.128713)=−40.624008)),(((9/(ln 2))×(−0.160858)=2.088621)) :})),(((9/(ln 2))W(10^(1/4) ×((ln 2)/9))=(9/(ln 2))×0.121311=1.575133)) :}

$$\mathrm{log}\:\left(\mathrm{4}^{\mathrm{2}{x}} ×{x}^{\mathrm{36}} \right)=\mathrm{9} \\ $$$$\mathrm{4}^{\mathrm{2}{x}} ×{x}^{\mathrm{36}} =\mathrm{10}^{\mathrm{9}} \\ $$$$\mathrm{2}^{\frac{{x}}{\mathrm{9}}} ×{x}=\pm\mathrm{10}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${e}^{\frac{{x}}{\mathrm{9}}\mathrm{ln}\:\mathrm{2}} ×{x}=\pm\mathrm{10}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${e}^{\frac{{x}}{\mathrm{9}}\mathrm{ln}\:\mathrm{2}} ×\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{9}}=\pm\mathrm{10}^{\frac{\mathrm{1}}{\mathrm{4}}} ×\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{9}} \\ $$$$\frac{{x}\mathrm{ln}\:\mathrm{2}}{\mathrm{9}}={W}\left(\pm\mathrm{10}^{\frac{\mathrm{1}}{\mathrm{4}}} ×\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{9}}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{9}}{\mathrm{ln}\:\mathrm{2}}{W}\left(\pm\mathrm{10}^{\frac{\mathrm{1}}{\mathrm{4}}} ×\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{9}}\right) \\ $$$$\:\:\:\:\:=\begin{cases}{\frac{\mathrm{9}}{\mathrm{ln}\:\mathrm{2}}{W}\left(−\mathrm{10}^{\frac{\mathrm{1}}{\mathrm{4}}} ×\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{9}}\right)=\begin{cases}{\frac{\mathrm{9}}{\mathrm{ln}\:\mathrm{2}}×\left(−\mathrm{3}.\mathrm{128713}\right)=−\mathrm{40}.\mathrm{624008}}\\{\frac{\mathrm{9}}{\mathrm{ln}\:\mathrm{2}}×\left(−\mathrm{0}.\mathrm{160858}\right)=\mathrm{2}.\mathrm{088621}}\end{cases}}\\{\frac{\mathrm{9}}{\mathrm{ln}\:\mathrm{2}}{W}\left(\mathrm{10}^{\frac{\mathrm{1}}{\mathrm{4}}} ×\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{9}}\right)=\frac{\mathrm{9}}{\mathrm{ln}\:\mathrm{2}}×\mathrm{0}.\mathrm{121311}=\mathrm{1}.\mathrm{575133}}\end{cases} \\ $$

Commented by Tawa11 last updated on 29/Aug/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by Gbenga last updated on 31/Aug/21

great sir

$${great}\:{sir} \\ $$

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