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) =(d/(fx+g)) (a,b,c,d,f,g∈R, a≠1,d≠0,f≠0,b≠0,a>0) ⇒(fx+g)e^((bx+c)lna) =d (f((blna)/(blna))x+g)e^(clna) e^(bxlna) =d ((fe^(clna−((gblna)/f)) )/(blna))(bxlna+((gblna)/f))e^(bxlna+((gblna)/f)) =d bxlna+((gblna)/f)=W{((dblna)/f)e^(((gblna)/f)−clna) } x=(1/(blna))[W{((dblna)/f)e^(((gblna)/f)−clna) }−((gblna)/f)] e=Euler′s constant In your problem, b=f=1,c=g=0,d=192,a=4 ⇒x=(1/(ln4))W{192ln4}](https://www.tinkutara.com/question/Q6871.png)
$${a}^{{bx}+{c}} =\frac{{d}}{{fx}+{g}}\:\:\:\:\:\left({a},{b},{c},{d},{f},{g}\in\mathbb{R},\:{a}\neq\mathrm{1},{d}\neq\mathrm{0},{f}\neq\mathrm{0},{b}\neq\mathrm{0},{a}>\mathrm{0}\right) \\ $$$$\Rightarrow\left({fx}+{g}\right){e}^{\left({bx}+{c}\right){lna}} ={d} \\ $$$$\left({f}\frac{{blna}}{{blna}}{x}+{g}\right){e}^{{clna}} {e}^{{bxlna}} ={d} \\ $$$$\frac{{fe}^{{clna}−\frac{{gblna}}{{f}}} }{{blna}}\left({bxlna}+\frac{{gblna}}{{f}}\right){e}^{{bxlna}+\frac{{gblna}}{{f}}} ={d} \\ $$$${bxlna}+\frac{{gblna}}{{f}}={W}\left\{\frac{{dblna}}{{f}}{e}^{\frac{{gblna}}{{f}}−{clna}} \right\} \\ $$$${x}=\frac{\mathrm{1}}{{blna}}\left[{W}\left\{\frac{{dblna}}{{f}}{e}^{\frac{{gblna}}{{f}}−{clna}} \right\}−\frac{{gblna}}{{f}}\right] \\ $$$${e}={Euler}'{s}\:{constant} \\ $$$${In}\:{your}\:{problem},\:{b}={f}=\mathrm{1},{c}={g}=\mathrm{0},{d}=\mathrm{192},{a}=\mathrm{4} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{{ln}\mathrm{4}}{W}\left\{\mathrm{192}{ln}\mathrm{4}\right\} \\ $$