Resposta:

g) S = {∛7}

h) S = {1/4}

i) S = {0, 2, 3}

Explicação passo a passo:

[tex]g)\: \log_x 7 = 3\to x^{3}=7\to x=\sqrt[3]{7}\\\\[/tex]

[tex]h)\: \log\sqrt{x}-\log x^{2} =\log \left(\dfrac{2}{x}\right) \to \log \left(x\right)^{\frac{1}{2}}-2\log x = \log 2 - \log x \to\\\\\\\dfrac{1}{2}\log x-2\log x + \log x= \log 2 \to \dfrac{1}{2}\log x-\log x= \log 2\to\\\\\\-\dfrac{1}{2}\log x= \log 2\to \log x= -2\log 2\to \log x=\log 2^{-2}\to\\\\\\x}=2^{-2}\to x=\dfrac{1}{2^{2}}\to x=\dfrac{1}{4}\\\\\\[/tex]

[tex]i)\: \log_{5x^{2}-6x} (8) = \log_x 2\to \dfrac{\log_3 8}{log_3 (5x^{2}-6x)} = \dfrac{\log_3 2}{\log_3 x}\to \\\\\\\dfrac{\log_3 8}{\log_3 2} = \dfrac{\log_3 (5x^{2}-6x)}{\log_3 x}\to \log_2 8 = \log_x (5x^{2}-6x)\to\\\\\\\log_2 2^{3} = \log_x (5x^{2}-6x)\to 3\log_2 2=\log_x (5x^{2}-6x)\to\\\\\\3\cdot 1 =\log_x (5x^{2}-6x)\to x^{3} = 5x^{2}-6x\to x^{3}-5x^{2}+6x=0\to \\\\x(x-2)(x-3)=0\\x=0\\x-2=0\to x = 2\\x-3=0\to x=3[/tex]

Espero ter ajudado! Qualquer dúvida pode perguntar nos comentários.