[tex]\boxed{\begin{matrix}\text{propriedades de logaritmos}: \\\\ \displaystyle \sf \log_{c^{n}}a =\frac{1}{n}\cdot \log_ca\\\\ \displaystyle \sf \log_ca^{n} = n \cdot \log_ca \\\\\ \displaystyle \sf \log_ca-\log_cb = \log_c\left(\frac{a}{b}\right) \end{matrix}}[/tex]

temos :

[tex]\displaystyle \sf \log_{0,125}\left(\frac{1}{h-8}\right)+\log_{\frac{1}{8}} (\sqrt{h}-2\sqrt{2}) = 8 \\\\\\ \log_{8^{-1}} (h-8)^{-1} +\log_{8^{-1}}(\sqrt{h} -2\sqrt{2}) = 8 \\\\\\ -1\cdot (-1)\cdot \log_8(h-8)-1\cdot \log_8(\sqrt{h}-2\sqrt{2}) = 8 \\\\ \log_8(h-8)-\log_8(\sqrt{h}-2\sqrt{2})=8 \\\\ \log_8\left(\frac{h-8}{\sqrt{h}-2\sqrt{2}}\right) = 8 \\\\\\[/tex]

[tex]\displaystyle \sf \frac{h-8}{\sqrt{h}-2\sqrt{2}} = 8^{8} \\\\\\ obs : \\\\ h-8 = \left[\sqrt{h}^2-(2\sqrt{2})^2\right] = (\sqrt{h}+2\sqrt{2})(\sqrt{h}-2\sqrt{2}) \\\\ Da{\'i}}: \\\\ \frac{(\sqrt{h}+2\sqrt{2})(\sqrt{h}-2\sqrt{2})}{\sqrt{h}-2\sqrt{2}}=8^{8} \\\\\\ \sqrt{h}+2\sqrt{2} = (2^{3})^{8} \\\\\\ \sqrt{h} = 2^{24}-2\sqrt{2} \\\\\\ \Large\boxed{\sf \ h = \left(2^{24}-2\sqrt{2}\right)^{2} \ }\checkmark[/tex]

Verified answer

Resposta:

Portanto,

[tex]h = (2 {}^{24} \: - 2 {}^{ \frac{3}{2} }) {}^{2}[/tex]

Explicação passo-a-passo:

[tex] log_{ \frac{1}{8} }( \frac{1}{h - 8} ) + log_{ \frac{1}{8} }( \sqrt{h} \: - \: 2 \sqrt{2}) = 8 \\ - log_{8}( \frac{ \sqrt{h} \: - \: 2 \sqrt{2} }{h - 8} ) = 8 \\ log_{8}( \frac{ \sqrt{h} \: - \: \sqrt{8} }{( \sqrt{h} \: - \: \sqrt{8} )( \sqrt{h} \: + \: \sqrt{8})}) \: = - 8 \\ log_{8}( \frac{1}{ \sqrt{h} \: + \: \sqrt{8} }) = - 8 \\ log_{8}(( \sqrt{h} \: + \: \sqrt{8} ) {}^{ - 1} ) = - 8 \\ - log_{8}( \sqrt{h} \: + \sqrt{8} ) = - 8 \\ log_{8}( \sqrt{h} \: + \sqrt{8} ) = 8 \\ \sqrt{h} \: + \sqrt{8} = 8 {}^{8} \\ \sqrt{h} = (2 {}^{3} ) {}^{8} \: - 2 {}^{ \frac{3}{2} } \\ h = (2 {}^{24} \: - 2 {}^{ \frac{3}{2} }) {}^{2}[/tex]