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Question 15 (5 points)

Express each logarithm in terms of [tex]\ln 3[/tex] and [tex]\ln 5[/tex].

[tex]\ln \frac{81}{125}[/tex]


Sagot :

To express \(\ln \frac{81}{125}\) in terms of \(\ln 3\) and \(\ln 5\), follow these steps:

1. Apply the properties of logarithms: Use the property that \(\ln \frac{a}{b} = \ln a - \ln b\).
[tex]\[ \ln \frac{81}{125} = \ln 81 - \ln 125 \][/tex]

2. Prime factorize each number:
- \(81\) can be written as \(3^4\).
- \(125\) can be written as \(5^3\).

3. Substitute these factorizations into the logarithms:
[tex]\[ \ln 81 = \ln (3^4) \][/tex]
[tex]\[ \ln 125 = \ln (5^3) \][/tex]

4. Apply the power rule of logarithms: The power rule states that \(\ln (a^b) = b \ln a\).
[tex]\[ \ln (3^4) = 4 \ln 3 \][/tex]
[tex]\[ \ln (5^3) = 3 \ln 5 \][/tex]

5. Combine the results: Substitute these back into the original expression.
[tex]\[ \ln \frac{81}{125} = \ln 81 - \ln 125 = 4 \ln 3 - 3 \ln 5 \][/tex]

So, the logarithm \(\ln \frac{81}{125}\) expressed in terms of \(\ln 3\) and \(\ln 5\) is:
[tex]\[ \ln \frac{81}{125} = 4 \ln 3 - 3 \ln 5 \][/tex]