Harper Clarke
03/02/2024 · Middle School
Suppose that \( \ln 3= \) a and \( \ln 11=b \). Use properties of logarithms to write the logarithm in terms of a and \( b \). \( \ln \sqrt[3]{33} \) \( \ln \sqrt[3]{33}=\square \)
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Step-by-step Solution
To express \( \ln \sqrt[3]{33} \) in terms of \( a \) and \( b \), we can start by rewriting \( 33 \) in terms of its prime factors:
\[
33 = 3 \times 11
\]
Now, we can use the properties of logarithms to rewrite \( \ln \sqrt[3]{33} \):
1. Use the property of logarithms that states \( \ln(x^n) = n \ln(x) \):
\[
\ln \sqrt[3]{33} = \ln(33^{1/3}) = \frac{1}{3} \ln(33)
\]
2. Next, we can express \( \ln(33) \) using the product property of logarithms, which states \( \ln(xy) = \ln(x) + \ln(y) \):
\[
\ln(33) = \ln(3 \times 11) = \ln(3) + \ln(11) = a + b
\]
3. Substitute this back into the expression for \( \ln \sqrt[3]{33} \):
\[
\ln \sqrt[3]{33} = \frac{1}{3} \ln(33) = \frac{1}{3}(a + b)
\]
Thus, we can write:
\[
\ln \sqrt[3]{33} = \frac{1}{3}(a + b)
\]
So the final answer is:
\[
\ln \sqrt[3]{33} = \frac{1}{3}(a + b)
\]
Quick Answer
\[
\ln \sqrt[3]{33} = \frac{1}{3}(a + b)
\]
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