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Question Number 120154 by mathace last updated on 29/Oct/20

solve using LambertW function  ((8/7))^x +17=25x

$${solve}\:{using}\:{LambertW}\:{function} \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{{x}} +\mathrm{17}=\mathrm{25}{x} \\ $$

Answered by mr W last updated on 29/Oct/20

((8/7))^x =25(x−((17)/(25)))  ((8/7))^(x−((17)/(25))) =25((8/7))^(−((17)/(25))) (x−((17)/(25)))  e^((x−((17)/(25)))ln (8/7)) =25((8/7))^(−((17)/(25))) (x−((17)/(25)))  −(1/(25))((8/7))^((17)/(25)) ln (8/7)=(((17)/(25))−x)(ln (8/7))e^((((17)/(25))−x)ln (8/7))   W[−(1/(25))((8/7))^((17)/(25)) ln (8/7)]=(((17)/(25))−x)ln (8/7)  ⇒x=((17)/(25))−(1/(ln (8/7)))W[−(1/(25))((8/7))^((17)/(25)) ln (8/7)]  ≈ { ((0.724057)),((53.864856)) :}

$$\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{{x}} =\mathrm{25}\left({x}−\frac{\mathrm{17}}{\mathrm{25}}\right) \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{{x}−\frac{\mathrm{17}}{\mathrm{25}}} =\mathrm{25}\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{−\frac{\mathrm{17}}{\mathrm{25}}} \left({x}−\frac{\mathrm{17}}{\mathrm{25}}\right) \\ $$$${e}^{\left({x}−\frac{\mathrm{17}}{\mathrm{25}}\right)\mathrm{ln}\:\frac{\mathrm{8}}{\mathrm{7}}} =\mathrm{25}\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{−\frac{\mathrm{17}}{\mathrm{25}}} \left({x}−\frac{\mathrm{17}}{\mathrm{25}}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{25}}\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{\frac{\mathrm{17}}{\mathrm{25}}} \mathrm{ln}\:\frac{\mathrm{8}}{\mathrm{7}}=\left(\frac{\mathrm{17}}{\mathrm{25}}−{x}\right)\left(\mathrm{ln}\:\frac{\mathrm{8}}{\mathrm{7}}\right){e}^{\left(\frac{\mathrm{17}}{\mathrm{25}}−{x}\right)\mathrm{ln}\:\frac{\mathrm{8}}{\mathrm{7}}} \\ $$$${W}\left[−\frac{\mathrm{1}}{\mathrm{25}}\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{\frac{\mathrm{17}}{\mathrm{25}}} \mathrm{ln}\:\frac{\mathrm{8}}{\mathrm{7}}\right]=\left(\frac{\mathrm{17}}{\mathrm{25}}−{x}\right)\mathrm{ln}\:\frac{\mathrm{8}}{\mathrm{7}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{17}}{\mathrm{25}}−\frac{\mathrm{1}}{\mathrm{ln}\:\frac{\mathrm{8}}{\mathrm{7}}}{W}\left[−\frac{\mathrm{1}}{\mathrm{25}}\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{\frac{\mathrm{17}}{\mathrm{25}}} \mathrm{ln}\:\frac{\mathrm{8}}{\mathrm{7}}\right] \\ $$$$\approx\begin{cases}{\mathrm{0}.\mathrm{724057}}\\{\mathrm{53}.\mathrm{864856}}\end{cases} \\ $$

Commented by mathace last updated on 30/Oct/20

Correct! Thank you sir.

$${Correct}!\:{Thank}\:{you}\:{sir}. \\ $$

Commented by mathace last updated on 30/Oct/20

Would you plz help to solve the following  by LambertW or other spacial function  ((8/7))^x +17^x =25x

$${Would}\:{you}\:{plz}\:{help}\:{to}\:{solve}\:{the}\:{following} \\ $$$${by}\:{LambertW}\:{or}\:{other}\:{spacial}\:{function} \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{{x}} +\mathrm{17}^{{x}} =\mathrm{25}{x} \\ $$

Commented by mr W last updated on 30/Oct/20

no chance!

$${no}\:{chance}! \\ $$

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