1)
![\displaystyle \bigg(\frac13\bigg)^{x+2}=\frac1{27}= \bigg(\frac13\bigg)^3\\\\x+2=3,\;x=1 \displaystyle \bigg(\frac13\bigg)^{x+2}=\frac1{27}= \bigg(\frac13\bigg)^3\\\\x+2=3,\;x=1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbigg%28%5Cfrac13%5Cbigg%29%5E%7Bx%2B2%7D%3D%5Cfrac1%7B27%7D%3D%20%5Cbigg%28%5Cfrac13%5Cbigg%29%5E3%5C%5C%5C%5Cx%2B2%3D3%2C%5C%3Bx%3D1)
2)
![\displaystyle \log_{\frac12}x=-1\\x=\bigg(\frac12\bigg)^{-1}=2 \displaystyle \log_{\frac12}x=-1\\x=\bigg(\frac12\bigg)^{-1}=2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clog_%7B%5Cfrac12%7Dx%3D-1%5C%5Cx%3D%5Cbigg%28%5Cfrac12%5Cbigg%29%5E%7B-1%7D%3D2)
3)
![\displaystyle 3^{x+1}-3^x=18\\3\cdot 3^x-3^x=18\\3^x=18:2=3^2\\x=2 \displaystyle 3^{x+1}-3^x=18\\3\cdot 3^x-3^x=18\\3^x=18:2=3^2\\x=2](https://tex.z-dn.net/?f=%5Cdisplaystyle%203%5E%7Bx%2B1%7D-3%5Ex%3D18%5C%5C3%5Ccdot%203%5Ex-3%5Ex%3D18%5C%5C3%5Ex%3D18%3A2%3D3%5E2%5C%5Cx%3D2)
4)
0\qquad \\2x+2>0\qquad \end{matrix}\quad \begin{Bmatrix}x(x-3)+(x-3)=0\\|x|>1\qquad \qquad \qquad \\x>-1\qquad \qquad \qquad \end{matrix}" alt="\displaystyle \log_5(x^2-1)=\log_5(2x+2)\\\\\begin{Bmatrix}x^2-1=2x+2\\x^2-1>0\qquad \\2x+2>0\qquad \end{matrix}\quad \begin{Bmatrix}x(x-3)+(x-3)=0\\|x|>1\qquad \qquad \qquad \\x>-1\qquad \qquad \qquad \end{matrix}" align="absmiddle" class="latex-formula">
(x-3)(x+1) = 0 ⇔ x=3 или x=-1
x=3 подходит под условие |x|>1 и x>-1
x=-1 не подходит под условие x>-1.
Ответ: 3.