Ryan Carrillo
07/18/2024 · Elementary School
Write the expression as a sum and/or difference of logarithms. Express powers as factors. \( \log _{7}(2401 \mathrm{x}) \) \( \log _{7}(2401 \mathrm{x})=\square \) (Type an exact answer in simplified form.)
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Step-by-step Solution
To express \( \log_{7}(2401 \mathrm{x}) \) as a sum and/or difference of logarithms, we can use the properties of logarithms:
1. The logarithm of a product: \( \log_b(mn) = \log_b(m) + \log_b(n) \)
2. The logarithm of a power: \( \log_b(m^k) = k \cdot \log_b(m) \)
First, we can break down \( 2401 \) into its prime factors. We know that:
\[
2401 = 7^4
\]
Now we can rewrite the expression:
\[
\log_{7}(2401 \mathrm{x}) = \log_{7}(7^4 \cdot x)
\]
Using the product property of logarithms:
\[
\log_{7}(7^4 \cdot x) = \log_{7}(7^4) + \log_{7}(x)
\]
Now, applying the power property of logarithms to \( \log_{7}(7^4) \):
\[
\log_{7}(7^4) = 4 \cdot \log_{7}(7) = 4 \cdot 1 = 4
\]
Thus, we can combine these results:
\[
\log_{7}(2401 \mathrm{x}) = 4 + \log_{7}(x)
\]
So the final expression is:
\[
\log_{7}(2401 \mathrm{x}) = 4 + \log_{7}(x)
\]
Quick Answer
\( \log_{7}(2401 \mathrm{x}) = 4 + \log_{7}(x) \)
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