To find the integral part of the logarithm of \( 0.00592 \), we first calculate \( \log_{10}(0.00592) \).
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Convert \( 0.00592 \) to Scientific Notation:
\[ 0.00592 = 5.92 \times 10^{-3} \]
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Use the Properties of Logarithms:
Using the logarithmic property that states \( \log(a \times b) = \log(a) + \log(b) \):
\[ \log_{10}(0.00592) = \log_{10}(5.92 \times 10^{-3}) = \log_{10}(5.92) + \log_{10}(10^{-3}) \]
We know \( \log_{10}(10^{-3}) = -3 \).
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Calculate \( \log_{10}(5.92) \):
The value of \( \log_{10}(5.92) \) is approximately \( 0.772 \) (you can check this with a calculator).
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Combine the Logarithms:
\[ \log_{10}(0.00592) \approx 0.772 - 3 = -2.228 \]
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Integral Part:
The integral (or floor) part of \( -2.228 \) is \( -3 \).
Thus, the integral part of \( \log_{10}(0.00592) \) is \( -3 \).