Question

Write each expression as a single logarithm.

$3{\log }_5\left(3x+1\right)-2{\log }_5\left(2x-1\right)-{\log }_{5}x$

Short answer

The result is${\log }_5\left[\frac{{\left(3x+1\right)}^3}{{\left(2x-1\right)}^{2}x}\right]$.

Step-by-step solution

Step 1. Given information.

Consider the given information.

$3{\log }_5\left(3x+1\right)-2{\log }_5\left(2x-1\right)-{\log }_{5}x$

Step 2. Calculation.

Write each expression as a single logarithm.

$\begin{array}{rcl}3{\log }_5\left(3x+1\right)-2{\log }_5\left(2x-1\right)-{\log }_{5}x& =& {\log }_5{\left(3x+1\right)}^3-{\log }_5{\left(2x-1\right)}^2-{\log }_{5}x\\ & =& {\log }_5{\left(3x+1\right)}^3-\left[{\log }_5{\left(2x-1\right)}^2+{\log }_{5}x\right]\\ & =& {\log }_5{\left(3x+1\right)}^3-{\log }_5{\left(2x-1\right)}^{2}x\\ & =& {\log }_5\frac{{\left(3x+1\right)}^3}{{\left(2x-1\right)}^{2}x}\end{array}$

Step 3. Conclusion.

The obtained result is${\log }_5\frac{{\left(3x+1\right)}^3}{{\left(2x-1\right)}^{2}x}$.