(a) Without using Mathematical tables, find x, given that \(6 \log (x + 4) = \log 64\)
(b) If \(U = {1, 2, 3,4, 5, 6, 7, 8, 9, 10}, X = {1, 2, 4, 6, 7, 8, 9}, Y = {1, 2, 3, 4, 7, 9}\) and \(Z = {2, 3, 4, 7, 9}\). What is \(X \cap Y \cap Z' \)?
(b) If \(U = {1, 2, 3,4, 5, 6, 7, 8, 9, 10}, X = {1, 2, 4, 6, 7, 8, 9}, Y = {1, 2, 3, 4, 7, 9}\) and \(Z = {2, 3, 4, 7, 9}\). What is \(X \cap Y \cap Z' \)?
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Correct Answer: Option n
Explanation:
(a) \(6 \log (x + 4) = \log 64\)
\(\implies (x + 4)^{6} = 64\)
\((x + 4)^{6} = 2^{6}\)
\(\implies x + 4 = 2 ; x = 2 - 4 = -2\)
\(\therefore x = -2\)
(b) \(U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\)
\(X = {1, 2, 4, 6, 7, 8, 9}; Y = {1, 2, 3, 4, 7, 9} ; Z = {2, 3, 4, 7,9}\)
\(X \cap Y = {1, 2, 4, 7, 9}\)
\({1, 2, 4, 7, 9} \cap Z' = {1, 2, 4, 7, 9} \cap {1, 5, 6, 8, 10}\)
= {1}
(a) \(6 \log (x + 4) = \log 64\)
\(\implies (x + 4)^{6} = 64\)
\((x + 4)^{6} = 2^{6}\)
\(\implies x + 4 = 2 ; x = 2 - 4 = -2\)
\(\therefore x = -2\)
(b) \(U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\)
\(X = {1, 2, 4, 6, 7, 8, 9}; Y = {1, 2, 3, 4, 7, 9} ; Z = {2, 3, 4, 7,9}\)
\(X \cap Y = {1, 2, 4, 7, 9}\)
\({1, 2, 4, 7, 9} \cap Z' = {1, 2, 4, 7, 9} \cap {1, 5, 6, 8, 10}\)
= {1}