Answer
Solution set = $\displaystyle \{\frac{5}{4}\}$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{ll}
x \gt 0, & 4x-1 \gt 0\\
& x \gt 1/4
\end{array}\right]$
$ x \gt \displaystyle \frac{1}{4} \qquad (*)$
This is the condition eventual solutions must satisfy.
$\log_{5}x+\log_{5}(4x-1)=1$
LHS: apply the product rule
RHS: write $1$ as $\log_{5}5$
$\log_{5}[x(4x-1)]=\log_{5}5\qquad $ ... if $\log_{b}M=\log_{b}N,$ then M=N
$ x(4x-1)=5$
$4x^{2}-x=5$
$ 4x^{2}-x-5=0\qquad $ ... use the quadratic formula
$ x=\displaystyle \frac{1\pm\sqrt{(-1)^{2}-4(4)(-5)}}{2\cdot 4}=\frac{1\pm 9}{8}$
$ x=\displaystyle \frac{10}{8}\quad $ or $\displaystyle \quad x=\frac{-8}{8}$
$ x=\displaystyle \frac{5}{4}\quad $ or $\quad x=-1$
The negative solution does not satisfy (*), so the
solution set = $\displaystyle \{\frac{5}{4}\}$
