Answer
$2\log_5 x$ + $\frac{3}{2}\log_5 (1-5x)$ - $\frac{1}{2}\log_5 (x^3-x)$
Work Step by Step
$Expand$ $the$ $logarithmic$ $expression:$
$\log_5 (\frac{x^2(1-5x)^{\frac{3}{2}}}{\sqrt {x^3-x}})$
Use the Second Law of logarithms
$\log_5 (\frac{x^2(1-5x)^{\frac{3}{2}}}{\sqrt {x^3-x}})$ = $\log_5 (x^2 (1-5x)^{\frac{3}{2}})$ - $\log_5 \sqrt {x^3 - x}$
Use the First Law of Logarithms for $\log_5 (x^2 (1-5x)^{\frac{3}{2}})$
$\log_5 (x^2 \times(1-5x)^{\frac{3}{2}})$ = $\log_5 x^2$ + $\log_5 (1-5x)^{\frac{3}{2}}$
$\log_5 x^2$ + $\log_5 (1-5x)^{\frac{3}{2}}$ - $\log_5 \sqrt{x^3-x}$
Rewrite the root to exponent form
$\log_5 x^2$ + $\log_5 (1-5x)^{\frac{3}{2}}$ - $\log_5 (x^3-x)^{\frac{1}{2}}$
Use the Third Law of Logarithms for the whole expression
$\log_5 x^2 = 2\log_5 x$
$\log_5 (1-5x)^{\frac{3}{2}}$ = $\frac{3}{2}\log_5 (1-5x)$
$\log_5 (x^3-x)^{\frac{1}{2}}$ = $\frac{1}{2}\log_5 (x^3-x)$
$2\log_5 x$ + $\frac{3}{2}\log_5 (1-5x)$ - $\frac{1}{2}\log_5 (x^3-x)$
