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Start by expanding \(\frac{3(2^{n+1}) - 4(2^{n-1})}{2^{n+1} - 2^n}\): \(\frac{3 \times 2^n \times 2^1 - 2^2 \times 2^n \times 2^{-1}}{2^n \times 2 - 2^n}\) NUMERATOR : 2\(^n\) ( 3\(^1\) X 2\(^1\) - 2\(^2\) X 2\(^{-1}\) ) --> 2\(^n\) ( 3 X 2 — 4 X \(\frac{1}{2}\) ) --> 2\(^n\) ( 6 - 2 ) --> 2\(^n\) (4) DENOMINATOR : 2\(^n\) ( 2\(^1\) - 1 ) --> 2\(^n\) ( 2 - 1) --> 2\(^n\) => \(\frac{2^n ( 4)}{2^n}\) = 4

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: \(\left(\frac{16}{81}\right)^{-\frac{3}{4}} \times \sqrt{\frac{100}{81}} = \frac{15}{4}\)

3y\(^2\) = 27y 3y\(^2\) - 27y = 0 3y(y - 9) = 0 3y = 0 or y - 9 = 0 y = 0 or y = 9 y = 0 or 9

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

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

(a) Total number of fruits = 16 limes + 14 apples = 30 (i) P(both are good limes) = \(\frac{10}{30} \times \frac{9}{29} = \frac{3}{29}\) (ii) P(both are bad fruits) = \(\frac{12}{30} \times \frac{11}{29} = \frac{22}{145}\) (iii) P(one is a good apple and the other a bad lime) = \((\frac{8}{30} \times \frac{6}{29}) + (\frac{6}{30} \times \frac{8}{29})\) = \(\frac{8}{145} + \frac{8}{145}\) = \(\frac{16}{145}\) Note : The arrangement is not specific. Hence, it can be a good apple first and then a bad lime or a bad lime first and then a good apple. (b) \(\log_{3} (4x + 1) - \log_{3} (3x - 5) = 2\) \(\log_{3} (\frac{4x + 1}{3x - 5}) = 2\) \((\frac{4x + 1}{3x - 5}) = 3^{2}\) \(\frac{4x + 1}{3x - 5} = 9 \implies 4x + 1 = 9(3x - 5)\) \(4x + 1 = 27x - 45 \implies 23x = 46\) \(x = 2\)

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

\(27^{(1+x)}\) = 9 \(3^{3(1+x)} = 3^2\) 3(1+x) = 2 3 + 3x = 2 3x = -1 x = \(\frac{-1}{3}\)

\(log_810\) = X = \(log_8{2 x 5}\) \(log_82\) + \(log_85\) = X Base 8 can be written as \(2^3\) \(log_82 = y\) therefore \(2 = 8^y\) \(y = \frac{1}{3}\) \(\frac{1}{3} = log_82\) taking \(\frac{1}{3}\) to the other side of the original equation \(log_85 = X-\frac{1}{3}\) explanation courtesy of Oluteyu and Ifechuks OR Given \(\log_8 10 = X\), we can express \(\log_8 5\) as follows: \(\log_8 5 = \log_8 \left( \frac{10}{2} \right) = \log_8 10 - \log_8 2\) Using the change of base formula, we find: \(\log_8 2 = \frac{\log 2}{\log 8} = \frac{1}{3}\) Thus, substituting back, we get: \(\log_8 5 = X - \log_8 2 = X - \frac{1}{3}\)

Start your solution by cross-multiplying, \(\frac{\sqrt{2}}{x+\sqrt{2}}=\frac{1}{x-\sqrt{2}}\) [x - \(\sqrt{2}\)]\(\sqrt{2}\) = x + \(\sqrt{2}\) where \(\sqrt{2}\) \times\(\sqrt{2}\) = 2 x\(\sqrt{2} - \sqrt{2} x \sqrt{2}\) = x + \(\sqrt{2}\) then collect like terms x\(\sqrt{2}\) - x = \(\sqrt{2}\) + 2 and factorize accordingly to get the unknown. x(\(\sqrt{2}\) - 1) = \(\sqrt{2}\) + 2 x = \(\frac{\sqrt{2}) + 2}{\sqrt{2} - 1}\) rationalize x = \(\frac{\sqrt{2} + 2}{\sqrt{2} - 1}\) * \(\frac{\sqrt{2} + 1}{\sqrt{2} + 1}\) x = \(\frac{\sqrt{4} + \sqrt{2} + 2\sqrt{2} + 2}{\sqrt{4} + \sqrt{2} - \sqrt{2} - 1}\) x = \(\frac{2 + 3\sqrt{2} + 2}{2 - 1}\) x = \(\frac{3\sqrt{2} + 4}{1}\) x = 3\(\sqrt{2}\) + 4

(2 x 9 \(^0\) ) (Y x 9 \(^1\)+ 3 x 9\(^0\)) = (3 x 5 \(^0\))(Y x 5\(^1\) + 3 x 5\(^0\)). (2 * 1)(Y *9 + 3) = (3 * 1)(Y * 5 + 3) 18y + 6 = 15y + 9 18y - 15y = 9 - 6 3y = 3 y = 1

Multiply appropriately to remove decimals on both numerator and denominators. Such that you have \(\sqrt{\frac{(23000 \times 750)}{(345 \times 125)}}\) Dividing, => \(\sqrt{400} = 20\).

Hint: Make a sketch forming a right angled triangle. Let x = height of the tree above the man. Such that x/25 = sin 30. x = 12.5m The height of the tree = 12.5 + 1.7 = 14.2 m

\DeltaABE and \DeltaACD are similar therefore \(\frac{ x}{ x + 4} = \frac{3}{6}\) 6x = 3(x+4) 6x = 3x + 12 3x = 12 x = 4 In \DeltaACD; \(L^2\) = \((4+4)^2 + 6^2\) = \(8^2 + 6^2\) = 64 + 36 = 100 L = \(\sqrt{100}\) = 10cm In ABE, \((AE)^2 = 4^2 + 3^2\) = 16 + 9 = 25 AE = 5cm ∴L = 10 - 5 = 5cm