Find the Exact Value log base 3 of 54- log base 3 of 8/15+ log base 3 of 4/5
The given problem is a logarithmic expression that involves finding the exact value of a combination of logs with the same base (base 3) using various properties of logarithms. Specifically, you are asked to perform three operations:
Find the logarithm base 3 of the number 54.
Subtract from that the logarithm base 3 of the fraction 8/15.
Add the logarithm base 3 of the fraction 4/5.
Understanding and solving this problem would involve applying the laws of logarithms, such as the quotient rule (log_b (x/y) = log_b(x) - log_b(y)), the product rule (log_b(x*y) = log_b(x) + log_b(y)), and possibly converting between exponential and logarithmic form if necessary. The goal is to simplify the expression to a form where the exact value can be derived without using a calculator.
$\left(log\right)_{3} \left(\right. 54 \left.\right) - \left(log\right)_{3} \left(\right. \frac{8}{15} \left.\right) + \left(log\right)_{3} \left(\right. \frac{4}{5} \left.\right)$
Answer
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 $\log_3\left(\frac{54}{\frac{8}{15}} \cdot \frac{4}{5}\right)$.
Step 3:
Multiply the numerator by the reciprocal of the denominator to simplify: $\log_3\left(54 \cdot \frac{15}{8} \cdot \frac{4}{5}\right)$.
Step 4:
Identify and cancel out common factors.
Step 4.1:
Extract the factor of 2 from 54: $\log_3\left(2 \cdot 27 \cdot \frac{15}{8} \cdot \frac{4}{5}\right)$.
Step 4.2:
Extract the factor of 2 from 8: $\log_3\left(2 \cdot 27 \cdot \frac{15}{2 \cdot 4} \cdot \frac{4}{5}\right)$.
Step 4.3:
Cancel out the common factor of 2: $\log_3\left(\cancel{2} \cdot 27 \cdot \frac{15}{\cancel{2} \cdot 4} \cdot \frac{4}{5}\right)$.
Step 4.4:
Rewrite the expression without the canceled factor: $\log_3\left(27 \cdot \frac{15}{4} \cdot \frac{4}{5}\right)$.
Step 5:
Combine 27 and $\frac{15}{4}$: $\log_3\left(\frac{27 \cdot 15}{4} \cdot \frac{4}{5}\right)$.
Step 6:
Multiply 27 by 15: $\log_3\left(\frac{405}{4} \cdot \frac{4}{5}\right)$.
Step 7:
Identify and cancel out common factors.
Step 7.1:
Factor out 5 from 405: $\log_3\left(\frac{5 \cdot 81}{4} \cdot \frac{4}{5}\right)$.
Step 7.2:
Cancel out the common factor of 5: $\log_3\left(\frac{\cancel{5} \cdot 81}{4} \cdot \frac{4}{\cancel{5}}\right)$.
Step 7.3:
Rewrite the expression without the canceled factor: $\log_3\left(\frac{81}{4} \cdot 4\right)$.
Step 8:
Cancel out the common factor of 4.
Step 8.1:
Cancel the common factor: $\log_3\left(\frac{81}{\cancel{4}} \cdot \cancel{4}\right)$.
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 $\log_3(81) = 4$.
