hence, we have
[tex]1=\frac{(-3x+5)}{2^x}[/tex]or equivalently
[tex]\begin{gathered} 1=(-3x+5)e^{-x\ln 2}\ldots\ldots.(A) \\ \end{gathered}[/tex]since
[tex]\frac{1}{2^x^{}}=2^{-x}=e^{-x\ln 2}[/tex]Now, from equation A, we can identify:
[tex]\begin{gathered} u=(-x+\frac{5}{3})\ln (2) \\ \text{and} \\ x=-\frac{3u-5\ln (2)}{3\ln (2)} \end{gathered}[/tex]by substituying these equations into A, we have,
[tex]1=(-3(-\frac{3u-5\ln2}{3\ln2})+5)e^{-\ln 2(-\frac{3u-5\ln 2}{3\ln 2})}[/tex]by simplifying this equation, we have
[tex]1=e^{\frac{3u-5\ln2}{3}}(\frac{3u-5\ln 2}{\ln 2}+5)[/tex]The final solution is
[tex]x=-\frac{3W(\frac{e^{5\ln2/3}\ln2}{3})-5\ln 2}{3\ln 2}[/tex]where W is given by
[tex]W=\frac{e^{-5\ln 2/3+2}}{e^{\frac{3u-5\ln 2+6}{5}}}[/tex]