$\begin{array}{l}1)\,\,4^{2x - 1} = 1\\ \Leftrightarrow 2x - 1 = 0\\ \Leftrightarrow x = \dfrac{1}{2}\\ 5)\,\,2^{x^2 -x + 8} = 4^{1 - 3x}\\ \Leftrightarrow 2^{x^2 -x + 8} = 2^{2 - 6x}\\ \Leftrightarrow x^2 - x +8 = 2 - 6x\\ \Leftrightarrow x^2 + 5x + 6 = 0\\ \Leftrightarrow \left[\begin{array}{l}x = -2\\x = -3\end{array}\right.\\ 6)\,\,2^{x^2 -6x - \frac{5}{2}}= 16\sqrt2\\ \Leftrightarrow 2^{x^2 -6x - \frac{5}{2}}= 2^4.2^{\frac{1}{2}} = 2^{\frac{9}{2}}\\ \Leftrightarrow x^2 -6x - \dfrac{5}{2} = \dfrac{9}{2}\\ \Leftrightarrow x^2 - 6x - 7 = 0\\ \Leftrightarrow \left[\begin{array}{l}x = -1\\x = 7\end{array}\right.\\ 3)\,\,5^x +5^{x+1} + 5^{x+2} = 3^x + 3^{x+3} - 3^{x+1}\\ \Leftrightarrow 5^x +5.5^x + 25.5^x = 3^x + 27.3^x - 3.3^x\\ \Leftrightarrow 31.5^x = 25.3^x\\ \Leftrightarrow \left(\dfrac{5}{3}\right)^x = \dfrac{25}{31}\\ \Leftrightarrow \ln\left(\dfrac{5}{3}\right)^x = \ln\dfrac{25}{31}\\ \Leftrightarrow x.\ln\dfrac{5}{3} = \ln\dfrac{25}{31}\\ \Leftrightarrow x = \dfrac{\ln\dfrac{25}{31}}{\ln\dfrac{5}{3}} \end{array}$
