Question

In the following exercises, use properties of logarithms to write each logarithm as a sum of logarithms. Simplify if possible.

${\log }_5\frac{8a^2{b}^{6}c}{d^3}$.

Short answer

The solution is$3{\log }_5\left(2\right)+2{\log }_5\left(a\right)+6{\log }_5\left(b\right)+{\log }_5\left(c\right)-3{\log }_5\left(d\right)$.

Step-by-step solution

Step 1. Given and explanation. 

We have ${\log }_5\frac{8a^2{b}^{6}c}{d^3}$.

The product property states that if there is an expression as ${\log }_a(m·n)$, it is equal to ${\log }_{a}m+{\log }_{a}n$.

The quotient property states that if there is an expression as ${\log }_a\frac{m}{n}$, it is equal to ${\log }_{a}m-{\log }_{a}n$.

Applying these properties, we get the solution.

Step 2. Expanding using quotient property.

We have ${\log }_5\frac{8a^2{b}^{6}c}{d^3}$.

Using the quotient property, we can write this as ,

${\log }_5\left(8a^2{b}^{6}c\right)-{\log }_5\left(d^3\right)$.

Step 3. Expanding using product property.

We have ${\log }_5\left(8a^2{b}^{6}c\right)-{\log }_5\left(d^3\right)$.

Using the product rule, we get
$$\begin{gathered} ={\log }_5\left(8a^2{b}^{6}c\right)-{\log }_5\left(d^3\right) \\ =\left[({\log }_{5}8)+({\log }_5{a}^2)+({\log }_5{b}^6)+({\log }_{5}c)\right]-{\log }_5\left(d^3\right) \\ ={\log }_5{\left(2\right)}^3+{\log }_5{\left(a\right)}^2+{\log }_5{\left(b\right)}^6+{\log }_{5}c-{\log }_5{\left(d\right)}^3=3{\log }_5\left(2\right)+2{\log }_5\left(a\right)+6{\log }_5\left(b\right)+{\log }_5\left(c\right)-3{\log }_5\left(d\right) \end{gathered}$$
Thus we get the value.