Question Number 6870 by Tawakalitu. last updated on 31/Jul/16 | ||
$${Solve}\:{for}\:{x}\:{in}\: \\ $$$$ \\ $$$$\mathrm{4}^{{x}} \:=\:\frac{\mathrm{192}}{{x}} \\ $$$$ \\ $$$${Using}\:{lambert}\:{function}\: \\ $$ | ||
Commented by Yozzii last updated on 31/Jul/16 | ||
$${a}^{{bx}+{c}} =\frac{{d}}{{fx}+{g}}\:\:\:\:\:\left({a},{b},{c},{d},{f},{g}\in\mathbb{R},\:{a}\neq\mathrm{1},{d}\neq\mathrm{0},{f}\neq\mathrm{0},{b}\neq\mathrm{0},{a}>\mathrm{0}\right) \\ $$$$\Rightarrow\left({fx}+{g}\right){e}^{\left({bx}+{c}\right){lna}} ={d} \\ $$$$\left({f}\frac{{blna}}{{blna}}{x}+{g}\right){e}^{{clna}} {e}^{{bxlna}} ={d} \\ $$$$\frac{{fe}^{{clna}−\frac{{gblna}}{{f}}} }{{blna}}\left({bxlna}+\frac{{gblna}}{{f}}\right){e}^{{bxlna}+\frac{{gblna}}{{f}}} ={d} \\ $$$${bxlna}+\frac{{gblna}}{{f}}={W}\left\{\frac{{dblna}}{{f}}{e}^{\frac{{gblna}}{{f}}−{clna}} \right\} \\ $$$${x}=\frac{\mathrm{1}}{{blna}}\left[{W}\left\{\frac{{dblna}}{{f}}{e}^{\frac{{gblna}}{{f}}−{clna}} \right\}−\frac{{gblna}}{{f}}\right] \\ $$$${e}={Euler}'{s}\:{constant} \\ $$$${In}\:{your}\:{problem},\:{b}={f}=\mathrm{1},{c}={g}=\mathrm{0},{d}=\mathrm{192},{a}=\mathrm{4} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{{ln}\mathrm{4}}{W}\left\{\mathrm{192}{ln}\mathrm{4}\right\} \\ $$ | ||