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104\(_x\) = 68 1 \(\times x^2 + 0 \times x + 4 \times x^0\) = 68 x\(^2\) = 68 - 4; x\(^2\) = 64 x = \(\sqrt{64}\) = 8 (104\(_x\) change to base 10)

5-3log52 x 22log23 i Let -3log52 = p => log52-3 = p ∴2-3 = 5p ∴5-3log52 = 5log52-3 = 5p ii 22log23 = q => log532 = q ∴32 = 2q ∴222log23 = 2q = 32 = 9 i x ii = 2-3 x 9 = 1/23 x 9 = 1/8 x 9 = 9/8 = 11/8

\(\left(\frac{16}{81}\right)^{-\frac{3}{4}}\times \sqrt{\frac{100}{81}}\) = \(\frac{1}{\left(\sqrt[4]{\frac{16}{81}}\right)^3}\times \frac{10}{9}\) = \(\frac{1}{\left(\frac{2}{3}\right)^3}\times\frac{10}{9}\) = \(\frac{27}{8}\times \frac{10}{9} = \frac{15}{4}\) OR We start with the expression: \(\left(\frac{16}{81}\right)^{-\frac{3}{4}} \times \sqrt{\frac{100}{81}}\) First, simplify \(\left(\frac{16}{81}\right)^{-\frac{3}{4}}\): \(\left(\frac{16}{81}\right)^{-\frac{3}{4}} = \left(\frac{81}{16}\right)^{\frac{3}{4}} = \frac{81^{\frac{3}{4}}}{16^{\frac{3}{4}}}\) Calculating \(81^{\frac{3}{4}}\) and \(16^{\frac{3}{4}}\): \(81 = 3^4 \implies 81^{\frac{3}{4}} = 3^3\) = 27 \(16 = 4^2 \implies 16^{\frac{3}{4}} = 4^{\frac{3}{2}} = 2^3\) = 8 Thus, \(\left(\frac{16}{81}\right)^{-\frac{3}{4}} = \frac{27}{8}\) Next, simplify \(\sqrt{\frac{100}{81}}\): \(\sqrt{\frac{100}{81}} = \frac{\sqrt{100}}{\sqrt{81}} = \frac{10}{9}\) Now, multiply the two results: \(\frac{27}{8} \times \frac{10}{9} = \frac{27 \times 10}{8 \times 9} = \frac{270}{72}\) Finding the GCD of 270 and 72 gives: \(\frac{270 \div 18}{72 \div 18} = \frac{15}{4}\) The final result is: \(\left(\frac{16}{81}\right)^{-\frac{3}{4}} \times \sqrt{\frac{100}{81}} = \frac{15}{4}\)

\(\log_4 x = -3\) \(x = 4^{-3}\) = \(\frac{1}{64}\)

Let the number = d. \(\log d = \bar{1}.8732\) \(\log \sqrt{d} = \frac{\bar{1}.8732}{2}\) = \(\frac{\bar{2} + 1.8732}{2}\) = \(\bar{1}.9366\) \(\therefore \sqrt{d} = Antilog (\bar{1}.9366)\) = 0.86

\(\frac{10}{\sqrt{32}}\) = \(\frac{10}{\sqrt{16\times 2}}\) = \(\frac{10}{4\sqrt{2}}\) \(\frac{10\times \sqrt{2}}{4\sqrt{2}\times\sqrt{2}}\) = \(\frac{10\sqrt{2}}{4\times 2}\) = \(\frac{5}{4}\sqrt{2}\)

By trial and error method 1728 is a perfect cube or try multiplying 3 same numbers based on the given options, or use the long division method, and group the divisors(same) in 3s. i.e., 12 x 12 x 12 = 1728, making 1728 a perfect cube.

(a) \(\frac{4.56 \times 3.6}{0.12} \equiv \frac{4.56 \times 360}{12}\) = \(4.56 \times 30 \) = \(136.8 \) = \(1.368 \times 10^{2}\) (b) \((73.8)^{2} - (26.2)^{2}\) Using the method of difference of two squares, = \((73.8 + 26.2) \times (73.8 - 26.2)\) = \(100 \times 47.6\) = \(4760\) (c) \(\sqrt{1\frac{19}{81}}\) = \(\sqrt{\frac{100}{81}}\) = \(\frac{10}{9}\)

\(\frac{7}{19}\) as a percentage will be \(\frac{7}{19}\times \frac{100}{1}\% = 36.8\%\)

398753 \(\approxeq\) 399000 (to 3 s.f)

\(2\frac{1}{6} + 2\frac{7}{12}\) = \(\frac{13}{6} + \frac{31}{12}\) = \(\frac{26 + 31}{12}\) = \(\frac{57}{12} = \frac{19}{4}\) \(\frac{19}{4} - 3\frac{1}{4}\) = \(\frac{19}{4} - \frac{13}{4}\) = \(\frac{6}{4}\) = \(1\frac{1}{2}\)

4321\(_7\) + 1234\(_7\) = 5555\(_7\) Missing number = 12341\(_7\) - 5555\(_7\) = 3453\(_7\)

Given that the ages are in the ratio 3: 4: 5. Let the sum of their ages be t. \(\therefore\) The youngest age = \(\frac{3}{12} t\) The eldest age = \(\frac{5}{12} t\) \(\implies \frac{5}{12} t - \frac{3}{12} t = 18\) \(\frac{2t}{12} = 18 \implies t = \frac{18 \times 12}{2}\) t = 108 years. The sum of their ages = 108 years.

Total capacity of the container = 45 liters = 45 x 1000 = 45000 cm\(^3\) Time to fill the container = \(\frac{45000}{2}\) = 22500 seconds = \(\frac{22500}{3600}\) = 6 hr 15 mins

Price of television = N25,000 Paid N10, 000 Remainder = N(25,000 - 10,000) = N15,000 Eight equal installments = \(\frac{15,000}{8}\) = N1,875 per installment.