Find the Exact Value log base 3 of 54- log base 3 of 8/15+ log base 3 of 4/5

Solution:

Step 1:

Apply the quotient rule of logarithms: ${\log }_b(x)-{\log }_b(y)={\log }_b\left(\frac{x}{y}\right)$. Thus, we get ${\log }_3\left(\frac{54}{\frac{8}{15}}\right)+{\log }_3\left(\frac{4}{5}\right)$.

Step 2:

Implement the product rule of logarithms: ${\log }_b(x)+{\log }_b(y)={\log }_b(xy)$. This gives us ${\log }_3(\frac{54}{\frac{8}{15}}\cdot \frac{4}{5})$.

Step 3:

Multiply the numerator by the reciprocal of the denominator to simplify: ${\log }_3(54\cdot \frac{15}{8}\cdot \frac{4}{5})$.

Step 4:

Identify and cancel out common factors.

Step 4.1:

Extract the factor of 2 from 54: ${\log }_3(2\cdot 27\cdot \frac{15}{8}\cdot \frac{4}{5})$.

Step 4.2:

Extract the factor of 2 from 8: ${\log }_3(2\cdot 27\cdot \frac{15}{2\cdot 4}\cdot \frac{4}{5})$.

Step 4.3:

Cancel out the common factor of 2: ${\log }_3(\cancel{2}\cdot 27\cdot \frac{15}{\cancel{2}\cdot 4}\cdot \frac{4}{5})$.

Step 4.4:

Rewrite the expression without the canceled factor: ${\log }_3(27\cdot \frac{15}{4}\cdot \frac{4}{5})$.

Step 5:

Combine 27 and $\frac{15}{4}$: ${\log }_3(\frac{27\cdot 15}{4}\cdot \frac{4}{5})$.

Step 6:

Multiply 27 by 15: ${\log }_3(\frac{405}{4}\cdot \frac{4}{5})$.

Step 7:

Identify and cancel out common factors.

Step 7.1:

Factor out 5 from 405: ${\log }_3(\frac{5\cdot 81}{4}\cdot \frac{4}{5})$.

Step 7.2:

Cancel out the common factor of 5: ${\log }_3(\frac{\cancel{5}\cdot 81}{4}\cdot \frac{4}{\cancel{5}})$.

Step 7.3:

Rewrite the expression without the canceled factor: ${\log }_3(\frac{81}{4}\cdot 4)$.

Step 8:

Cancel out the common factor of 4.

Step 8.1:

Cancel the common factor: ${\log }_3(\frac{81}{\cancel{4}}\cdot \cancel{4})$.

Step 8.2:

Rewrite the expression without the canceled factor: ${\log }_3(81)$.

Step 9:

Evaluate the logarithm base 3 of 81, which is 4: $4$.

Knowledge Notes:

• The quotient rule of logarithms states that the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator: ${\log }_b\left(\frac{x}{y}\right)={\log }_b(x)-{\log }_b(y)$.

• The product rule of logarithms states that the logarithm of a product is equal to the sum of the logarithms of the factors: ${\log }_b(xy)={\log }_b(x)+{\log }_b(y)$.

• When simplifying expressions involving logarithms, it's often helpful to factor out and cancel common factors to simplify the argument of the logarithm.

• The logarithm ${\log }_b(b^x)=x$ because $b^x$ is the inverse operation of ${\log }_b(x)$.

• In this problem, we used the fact that $3^4=81$ to conclude that ${\log }_3(81)=4$.