[tex]log_{ \frac{1}{4} } (7x+1) = log_9 27[/tex], D = ([tex] \frac{-1}{7} [/tex]; +∞)
[tex]a = log_b (b^a) =\ \textgreater \ log_9 27 = log_ \frac{1}{4} ( (\frac{1}{4})^{log_9 27})[/tex]
[tex]log_9 27 = \frac{3}{2} [/tex]
[tex] \frac{1}{4}^{ \frac{3}{2}} = \frac{1}{ \sqrt[]{2^6} } = \frac{1}{8} [/tex]
[tex]log_a f(x) = log_a g(x) =\ \textgreater \ f(x)=g(x)[/tex]
[tex]7x+1 = \frac{1}{8} =\ \textgreater \ x = \frac{-1}{8} [/tex] ∈ D
