(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 n...
Question 1 Report
(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.
Answer Details
(a) Write \(3.6 = \dfrac{36}{10} = \dfrac{2^2 \times 3^2}{10}\):
\[\log 3.6 = 2\log 2 + 2\log 3 - \log 10\]
\[= 2(0.3010) + 2(0.4771) - 1 = 0.6020 + 0.9542 - 1 = 0.5562\]
(b) All numbers are in base two. Convert to base ten: \(101_2 = 5\), \(11_2 = 3\), \(10010_2 = 18\). The equation \(\dfrac{y}{y + 101} = \dfrac{11}{10010}\) becomes
\[\frac{y}{y + 5} = \frac{3}{18} = \frac16\]
\[6y = y + 5 \;\Rightarrow\; 5y = 5 \;\Rightarrow\; y = 1_{10}\]
Converting back to base two, \(y = 1_2\).
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