Question Number 76176 by Master last updated on 24/Dec/19
Commented by mr W last updated on 24/Dec/19
![(3x+1)8^x =6 (3x+1)2^(3x) =6 (3x+1)2^(3x+1) =12 (3x+1)e^((3x+1)ln 2) =12 (3x+1)ln 2 e^((3x+1)ln 2) =12 ln 2 (3x+1)ln 2 =W(12 ln 2) ⇒x=(1/3)[((W(12 ln 2))/(ln 2))−1] ≈(1/3)[((1.629884547)/(ln 2))−1]=0.450475](https://www.tinkutara.com/question/Q76177.png)
$$\left(\mathrm{3}{x}+\mathrm{1}\right)\mathrm{8}^{{x}} =\mathrm{6} \\ $$$$\left(\mathrm{3}{x}+\mathrm{1}\right)\mathrm{2}^{\mathrm{3}{x}} =\mathrm{6} \\ $$$$\left(\mathrm{3}{x}+\mathrm{1}\right)\mathrm{2}^{\mathrm{3}{x}+\mathrm{1}} =\mathrm{12} \\ $$$$\left(\mathrm{3}{x}+\mathrm{1}\right){e}^{\left(\mathrm{3}{x}+\mathrm{1}\right)\mathrm{ln}\:\mathrm{2}} =\mathrm{12} \\ $$$$\left(\mathrm{3}{x}+\mathrm{1}\right)\mathrm{ln}\:\mathrm{2}\:{e}^{\left(\mathrm{3}{x}+\mathrm{1}\right)\mathrm{ln}\:\mathrm{2}} =\mathrm{12}\:\mathrm{ln}\:\mathrm{2} \\ $$$$\left(\mathrm{3}{x}+\mathrm{1}\right)\mathrm{ln}\:\mathrm{2}\:={W}\left(\mathrm{12}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{3}}\left[\frac{{W}\left(\mathrm{12}\:\mathrm{ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}}−\mathrm{1}\right] \\ $$$$\approx\frac{\mathrm{1}}{\mathrm{3}}\left[\frac{\mathrm{1}.\mathrm{629884547}}{\mathrm{ln}\:\mathrm{2}}−\mathrm{1}\right]=\mathrm{0}.\mathrm{450475} \\ $$
Commented by Master last updated on 24/Dec/19
$$\mathrm{thanks} \\ $$
Commented by benjo last updated on 25/Dec/19
$$\mathrm{sir}\:\mathrm{how}\:\mathrm{to}\:\mathrm{find}\:\mathrm{value}\:\mathrm{of}\:\mathrm{W}\left(\mathrm{12ln12}\right)?\: \\ $$$$\mathrm{please}\:\mathrm{write}\:\mathrm{the}\:\mathrm{formula} \\ $$
Commented by mr W last updated on 25/Dec/19
$${W}\:{function}'{s}\:{value}\:{can}\:{only}\:{be} \\ $$$${approximated},\:{see}\:{wikipedia}\:{for} \\ $$$${more}\:{about}\:{numerical}\:{evaluation}. \\ $$