0\\ 5-4^x=0\Rightarrow4^x=5\Rightarrow x=\log_45\approx1,16\\ x\in(-\infty;1,16):x=0\Rightarrow5-4^0=4>0\\ x\in(1,16;+\infty):x=2\Rightarrow5-4^2=-9<0\\ 2)\quad x+2>0\Rightarrow x>-2\\ \Righarrow x\in(-2;1,16)\\ \log_{x+2}\log_4(5-4^x)=1\\ (x+2)^1=\log_4(5-4^x)\\ 5-4^x = 4^{x+2}\\ 5-4^x=4^2\cdot4^x\\ 4^x(16+1)=5 4^x=\frac{5}{17}\\ x=\log_4\frac{5}{17}\approx-0,88\\ -0,88\in O.O.\Phi\\x\in(-2;-0,88)\cup(-0,88;1,16)" alt="\\ \log_{x+2}\log_4(5-4^x)=1\\ O.O.\Phi.:\\ 1)\quad 5-4^x>0\\ 5-4^x=0\Rightarrow4^x=5\Rightarrow x=\log_45\approx1,16\\ x\in(-\infty;1,16):x=0\Rightarrow5-4^0=4>0\\ x\in(1,16;+\infty):x=2\Rightarrow5-4^2=-9<0\\ 2)\quad x+2>0\Rightarrow x>-2\\ \Righarrow x\in(-2;1,16)\\ \log_{x+2}\log_4(5-4^x)=1\\ (x+2)^1=\log_4(5-4^x)\\ 5-4^x = 4^{x+2}\\ 5-4^x=4^2\cdot4^x\\ 4^x(16+1)=5 4^x=\frac{5}{17}\\ x=\log_4\frac{5}{17}\approx-0,88\\ -0,88\in O.O.\Phi\\x\in(-2;-0,88)\cup(-0,88;1,16)" align="absmiddle" class="latex-formula">
Интервалы проверьте сами.