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<p>For the given logarithmic expression:</p> <p>\(\log 7 + \log 2 + \log \frac{1}{40} + \log \frac{1}{7}\)</p> <p>We can simplify using the properties of logarithms:</p> <p>Combine \(\log 7\) and \(\log \frac{1}{7}\) using \(\log a + \log \frac{1}{a} = \log 1 = 0\):</p> <p>\(\log 7 + \log \frac{1}{7} = \log 1 = 0\)</p> <p>Combine \(\log 2\) and \(\log \frac{1}{40}\) by rewriting \(\frac{1}{40}\) as \(2^{-3}\), then use the property \(\log a^b = b \log a\):</p> <p>\(\log 2 + \log 2^{-3} = \log 2 + (-3)\log 2 = \log 2 - 3 \log 2\)</p> <p>\(\log 2 - 3 \log 2 = -2 \log 2\)</p> <p>Since \(\log 7 + \log \frac{1}{7} = 0\), all that's left is:</p> <p>\(-2 \log 2\)</p>
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