\[\boxed{\mathbf{39.}}\]
\[\textbf{а)}\log_{3}\sqrt{x} + \log_{3}{\sqrt{x + 8} =}1\]
\[\log_{3}{\sqrt{x}\sqrt{x + 8}} = \log_{3}3\]
\[x > 0;\]
\[\sqrt{x}\sqrt{x + 8} = 3\]
\[x(x + 8) = 9\]
\[x^{2} + 8x - 9 = 0\]
\[D_{1} = 16 + 9 = 25\]
\[x_{1} = - 4 + 5 = 1;\]
\[x_{2} = - 4 - 5 =\]
\[= - 9 < 0\ (не\ корень).\]
\[Ответ:x = 1.\]
\[\textbf{б)}\log_{2}\sqrt{x + 3} + \log_{2}\sqrt{x + 6} = 1\]
\[\log_{2}{\sqrt{x + 3}\sqrt{x + 6}} = \log_{2}2\]
\[x > - 3.\]
\[\sqrt{x + 3}\sqrt{x + 6} = 2\]
\[(x + 3)(x + 6) = 4\]
\[x^{2} + 3x + 6x + 18 - 4 = 0\]
\[x^{2} + 9x + 14 = 0\]
\[x_{1} + x_{2} = - 9;\ \ x_{1} \cdot x_{2} = 14\]
\[x_{1} = - 7 < - 3\ (не\ корень);\ \ \]
\[x + 2 = - 2.\]
\[Ответ:x = - 2.\]
\[\textbf{в)}\log_{3}\sqrt{x + 4} + \log_{3}\sqrt{x - 4} = 1\]
\[\log_{3}{\sqrt{x + 4}\sqrt{x - 4}} = \log_{3}3\]
\[x > 4.\]
\[\sqrt{x + 4}\sqrt{x - 4} = 3\]
\[(x + 4)(x - 4) = 9\]
\[x^{2} - 16 = 9\]
\[x^{2} = 25\]
\[x_{1} = - 5 < 4\ (не\ корень);\]
\[x_{2} = 5.\]
\[Ответ:x = 5.\]
\[\textbf{г)}\log_{4}\sqrt{x + 3} + \log_{4}\sqrt{x - 3} = 1\]
\[\log_{4}{\sqrt{x + 3}\sqrt{x - 3}} = \log_{4}4\]
\[x > 3.\]
\[\sqrt{x + 3}\sqrt{x - 3} = 4\]
\[(x + 3)(x - 3) = 16\]
\[x^{2} - 9 = 16\]
\[x^{2} = 25\]
\[x_{1} = - 5 < 3\ (не\ корень);\]
\[x_{2} = 5.\]
\[Ответ:x = 5.\]