(a) Evaluate without using the mathematical table or calculator, \(\log_{10} \sqrt{30} - \log_{10} \sqrt{6} + \log_{10} \sqrt{2}\). (b) \(U = {1, 2, 3, ...,...

Answer Details

(a) Evaluate \(\log_{10}\sqrt{30} - \log_{10}\sqrt{6} + \log_{10}\sqrt{2}\).

Write each surd as a power and combine using the laws of logarithms. Since \(\sqrt{n}=n^{1/2}\),

\[\tfrac{1}{2}\log_{10}30 - \tfrac{1}{2}\log_{10}6 + \tfrac{1}{2}\log_{10}2 = \tfrac{1}{2}\left(\log_{10}30 - \log_{10}6 + \log_{10}2\right).\]

Using \(\log a - \log b + \log c = \log\dfrac{ac}{b}\):

\[\tfrac{1}{2}\log_{10}\!\left(\frac{30\times 2}{6}\right) = \tfrac{1}{2}\log_{10}\!\left(\frac{60}{6}\right) = \tfrac{1}{2}\log_{10}10.\]

Since \(\log_{10}10 = 1\),

\[= \tfrac{1}{2}\times 1 = \boxed{\tfrac{1}{2}}.\]

(b) \(U=\{1,2,3,\dots,10\}\), \(A=\{1,2,3,4,5\}\), \(B=\{2,3,5\}\), \(C=\{6,8,10\}\).

(i) Filling the Venn diagram. Every element of \(B\) is also in \(A\) (\(2,3,5\in A\)), so the circle \(B\) lies completely inside circle \(A\), exactly as shown. \(C\) has no element in common with \(A\), so it is drawn separately. The regions are filled as:

• Inside \(B\) (which is inside \(A\)): \(2,\;3,\;5\)
• Inside \(A\) but outside \(B\): \(1,\;4\)
• Inside \(C\): \(6,\;8,\;10\)
• Outside every circle: \(7,\;9\)

(The diagram already shows \(1\) in the \(A\)-only region, \(5\) in \(B\), \(6,10\) in \(C\) and \(9\) outside; the remaining elements \(2,3\) join \(B\), \(4\) joins the \(A\)-only region, \(8\) joins \(C\) and \(7\) joins the outside.)

(ii) \(A\cap C\). \(A=\{1,2,3,4,5\}\) and \(C=\{6,8,10\}\) share no element, so

\[A\cap C = \varnothing\;\;(\text{the empty set}).\]

(iii) \(A\cap B'\). First \(B' = U - B = \{1,4,6,7,8,9,10\}\). Then take the elements common to \(A\):

\[A\cap B' = \{1,2,3,4,5\}\cap\{1,4,6,7,8,9,10\} = \{1,\;4\}.\]