Clark Hobbs
10/18/2023 · High School
Write the expression as a sum and/or difference of logarithms. Express powers as
factors.
\( \ln \left(x^{2} \sqrt{5-x}\right), 0
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Tutor-Verified Answer
Step-by-step Solution
To express \( \ln \left(x^{2} \sqrt{5-x}\right) \) as a sum and/or difference of logarithms, we can use the properties of logarithms:
1. The logarithm of a product: \( \ln(a \cdot b) = \ln(a) + \ln(b) \)
2. The logarithm of a power: \( \ln(a^b) = b \cdot \ln(a) \)
3. The logarithm of a square root: \( \ln(\sqrt{a}) = \frac{1}{2} \ln(a) \)
Now, let's apply these properties step by step:
1. Start with the expression:
\[
\ln \left(x^{2} \sqrt{5-x}\right)
\]
2. Use the product property of logarithms:
\[
\ln \left(x^{2} \sqrt{5-x}\right) = \ln(x^{2}) + \ln(\sqrt{5-x})
\]
3. Now, apply the power property to \( \ln(x^{2}) \):
\[
\ln(x^{2}) = 2 \ln(x)
\]
4. Next, apply the square root property to \( \ln(\sqrt{5-x}) \):
\[
\ln(\sqrt{5-x}) = \frac{1}{2} \ln(5-x)
\]
5. Combine these results:
\[
\ln \left(x^{2} \sqrt{5-x}\right) = 2 \ln(x) + \frac{1}{2} \ln(5-x)
\]
Thus, the final expression is:
\[
\ln \left(x^{2} \sqrt{5-x}\right) = 2 \ln(x) + \frac{1}{2} \ln(5-x)
\]
Quick Answer
\( \ln \left(x^{2} \sqrt{5-x}\right) = 2 \ln(x) + \frac{1}{2} \ln(5-x) \)
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