Answer:
[tex]\log_3(A) = -\dfrac{13}{28}\log_3(2) + \dfrac{7}{12}\log_3(7) - \dfrac12 \log_3(5) - \dfrac{1}7\log_3(37)[/tex]
Step-by-step explanation:
We are simplifying the expression [tex]\log_3(A)[/tex] where:
[tex]A= \dfrac{14^\frac{1}{4} \cdot \sqrt[3]{7}}{\sqrt{20}\cdot\sqrt[7]{148}}[/tex]
First, we can apply the division property of logarithms:
[tex]\log\!\left(\dfrac{a}{b}\right)=\log(a) - \log(b)[/tex]
↓↓↓
[tex]\log_3(A) = \log_3(14^\frac{1}{4} \cdot \sqrt[3]{7}) - \log_3(\sqrt{20}\sqrt[7]{148})[/tex]
Then, we can apply the multiplication property of logs:
[tex]\log(a\cdot b) = \log(a)+\log(b)[/tex]
↓↓↓
[tex]\log_3(A) = \log_3(14^\frac{1}{4}) + \log_3(\sqrt[3]{7}) - \left( \dfrac{}{}\log_3(\sqrt{20}) + \log_3(\sqrt[7]{148}) \dfrac{}{}\right)[/tex]
Remember to distribute the negative:
[tex]\log_3(A) = \log_3(14^\frac{1}{4}) + \log_3(\sqrt[3]{7}) - \log_3(\sqrt{20}) - \log_3(\sqrt[7]{148})[/tex]
Before proceeding, we can make this expression easier to work with by rewriting the radicals as exponents:
[tex]\sqrt[a]{x} = x^\frac{1}{a}[/tex]
↓↓↓
[tex]\log_3(A) = \log_3(14^\frac{1}{4}) + \log_3(7^\frac{1}{3}) - \log_3(20^\frac{1}{2}) - \log_3(148^\frac{1}{7})[/tex]
Next, we can apply the exponent property of logs:
[tex]\log(x^a) = a\log(x)[/tex]
↓↓↓
[tex]\log_3(A) = \dfrac{1}4\log_3(14) + \dfrac{1}{3}\log_3(7) - \dfrac{1}2\log_3(20) - \dfrac{1}7\log_3(148)[/tex]
Finally, we can rewrite the non-prime numbers (7, 20, and 148) into the product of their prime factors so that we can again apply the multiplication property:
[tex]\log_3(A) = \dfrac{1}4\log_3(2\cdot 7) + \dfrac{1}{3}\log_3(7) - \dfrac{1}2\log_3(2^2 \cdot 5) - \dfrac{1}7\log_3(2^2\cdot 37)[/tex]
[tex]= \dfrac{1}4\log_3(2) + \dfrac{1}4\log_3(7) + \dfrac{1}{3}\log_3(7) - 2\cdot \dfrac{1}2\log_3(2) - \dfrac12 \log_3(5) - 2\cdot\dfrac{1}7\log_3(2) - \dfrac{1}7\log_3(37)[/tex]
And now we can combine like terms:
[tex]\boxed{\log_3(A) = -\dfrac{13}{28}\log_3(2) + \dfrac{7}{12}\log_3(7) - \dfrac12 \log_3(5) - \dfrac{1}7\log_3(37)}[/tex]
