Answer
The solution set is $\{\frac{5}{4}\}$.
Work Step by Step
The given equation is
$\log _5 x+\log _5(4x-1)=1$
Use product rule.
$\log _5 [x(4x-1)]=1$
Rewrite in exponential form.
$x(4x-1)= 5^1$
Use the distributive property on the left.
$4x^2-x=5$
Subtract $5$ from both sides.
$4x^2-x-5=5-5$
Simplify.
$4x^2-x-5=0$
Factor.
$4x^2-5x+4x-5=0$
$x(4x-5)+1(4x-5)=0$
$(4x-5)(x+1)=0$
Set both factors equal to $0$.
$4x-5=0$ or $x+1=0$
Isolate $x$.
$x=\frac{5}{4}$ or $x=-1$
$-1$ is not in the domain of a logarithmic function.
The solution is $x=\frac{5}{4}$.
