Question

$log_{ \frac{1}{4} }(7x+1)= log_{9} 27$

Answer 1

$log_{ \frac{1}{4} }(7x+1)= log_{ 3^{2}} 3^3 \\ log_{ \frac{1}{4} }(7x+1)= \frac{3}{2} \\ x\ \textgreater \ - \frac{1}{7} \\ 7x+1=( \frac{1}4}) ^{ \frac{3}{2} } \\ 7x+1= \frac{1}{8} \\ 7x=- \frac{7}{8} \\ x=- \frac{1}{8}$

Answer 2

$\displaystyle \mathtt{log_{ \frac{1}{4} } (7x+1)=log_927~~~~~~~~~~~~~~~~~~~~~~~~~~C.E.~7x+1\ \textgreater \ 0 \Rightarrow x\ \textgreater \ - \frac{1}{7} } \\ \\ \mathtt{log_{ \frac{1}{4} }(7x+1)=log_{3^2}3^3} \\ \\ \mathtt{log_{ \frac{1}{4}}(7x+1)= \frac{1}{2}\cdot 3log_33 }\\ \\ \mathtt{log_{ \frac{1}{4} }(7x+1)= \frac{3}{2} }\\ \\ \mathtt{7x+1=\left( \frac{1}{4} \right)^{ \frac{3}{2}}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\mathtt{4^{ \frac{3}{2}}=\left(2^2\right)^{ \frac{3}{2}}}=2^{ \frac{6}{2} }=2^3=8}\\ \\ \mathtt{7x+1= \frac{1}{8} }$

$\displaystyle \mathtt{56x+8=1} \\ \\ \mathtt{56x=1-8} \\ \\ \mathtt{56x=-7|:7} \\ \\ \mathtt{ 8x=-1} \\ \\ \mathtt{x=- \frac{1}{8} }$