(a) Without using calculator or tables, find the value of \(\log 3.6\) given that \(\log 2 = 0.3010, \log 3 = 0.4771\) and \(\log 5 = 0.6990\).
(b) If all numbers in the equation \(\frac{y}{y + 101} = \frac{11}{10010}\) are in base two, solve for y.
(b) If all numbers in the equation \(\frac{y}{y + 101} = \frac{11}{10010}\) are in base two, solve for y.
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Correct Answer: Option n
Explanation:
(a) \(\log 3.6 = \log (\frac{18}{5})\)
\(\log 18 - \log 5 = \log (2 \times 3^{2}) - \log 5\)
\(\log 2 + \log 3^{2} - \log 5 = \log 2 + 2\log 3 - \log 5\)
= \(0.3010 + 2(0.4771) - 0.6990\)
= \(0.3010 + 0.9542 - 0.6990\)
= \(0.5562\).
(b) \(\frac{y}{y + 101} = \frac{11}{10010}\) (all in base 2)
Cross multiplying,
\(11(y + 101) = 10010y\)
\(11y + 1111 = 10010y\)
\(10010y - 11y = 1111\)
\(1111y = 1111\)
\(y = 1\).
(a) \(\log 3.6 = \log (\frac{18}{5})\)
\(\log 18 - \log 5 = \log (2 \times 3^{2}) - \log 5\)
\(\log 2 + \log 3^{2} - \log 5 = \log 2 + 2\log 3 - \log 5\)
= \(0.3010 + 2(0.4771) - 0.6990\)
= \(0.3010 + 0.9542 - 0.6990\)
= \(0.5562\).
(b) \(\frac{y}{y + 101} = \frac{11}{10010}\) (all in base 2)
Cross multiplying,
\(11(y + 101) = 10010y\)
\(11y + 1111 = 10010y\)
\(10010y - 11y = 1111\)
\(1111y = 1111\)
\(y = 1\).