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\( [ 1 ÷ 64^{(x + 2)}]= [4^{(x − 3)} ÷ 16^x ] \) \( 64^{−(x + 2)} = [4^{(x − 3)}] ÷[16^x] \) breakdown 4,16,64 into a small index no \( 2^{−6(x + 2)} = 2^{2(x − 3)} ÷ 2^{4(x)} \) \( 2^{−6x− 12} = 2^{2x − 4x − 6} \) \( 2^{−6x −12} = 2^{−2x − 6} \) − 6x − 12 = − 2x − 6 Collect the like term −6x + 2x = −6 + 12 −4x =6 x = \( \frac{6}{4} \) x = \( \frac{−3}{2} \)

log717 = [log 17 ÷ log7] = [1.2304 ÷ 0.8451] [100.0899 ÷ 101.9270] = 1.455(antilog)

\(\log_{5}(y^{2} x^{5} \div 125b^{\frac{1}{2}})\) = \(\log_{5} y^{2} + \log_{5} x^{5} - [\log_{5} 125 + \log_{5} b^{\frac{1}{2}}\) = \(2\log_{5} y + 5\log_{5} x - \log_{5} 5^{3} - \frac{1}{2} \log_{5} b\) = \(2\log_{5} y + 5\log_{5} x - 3 - \frac{1}{2}\log_{5} b\)

\(27^x ÷ 81^{(x + 2y)} = 9 \\ (27)x = 9 × 81^{(x+2y)} \\ (3^3 )^x =32 \times 3^{4(x + 2y)} \\ =3^{(2 + 4x + 8y)}\\ 3^{3x} = 3^{ (2 + 4x + 8y)}\\ 3x = 2 + 4x + 8y\\ 3x − 4x − 8y = 2 … … … (1)\\ x + 4y = 0 … … … (2)\\ − 4y = 2\\ y = (− 2) ÷ 4 = − ½\\ y = − ½\\ \) Substitute the value of y into equation (2) i.e x + 4y = 0 x + 4( − 1/2) = 0 x − 2 = 0 x = 2 ∴ x = 2,y = − ½) Method II \( 27^x ÷ 31^{(x + 2y) }= 9\\ 3^{3x} × 3^{( − 4x − 8y)} = 32\\ 3^{(3x − 8y)} = 32\\ − x − 8y=2 ……… (1)\\ x + 4y = 0 ……… (2)\\ − 4 = 2\\ y= 2/4 = ½\\ y = ½ \) Substitute the value of y into equation 2 x + 4y=0 x + 4 (− 1) ÷ 2) = 0 x − 2 = 0 x = 2 x = 2, y = ½

\(\sqrt{30}\) \times \(\sqrt{40}\) = \(\sqrt{(30 \times 40)}\) \(\sqrt{(3 \times 10 \times 4 \times 10)}\) \(\sqrt{(400 \times 3)}\) \(\sqrt{400}\) \times \(\sqrt{3}\) 20 \times \(\sqrt{3}\) 20\(\sqrt{3}\)

5÷ (2 − √3) Using conjugate surds [5 ÷ (2 − √3)]×(2 + √3) ÷ (2 + √3)] [5(2+√3)÷((2− √3))2] [5(2 +√3) ÷ (2)2− (√3)2] [5(2 +√3) ÷(4− 3)] = 5(2 + √3)

\(\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1\) \(\frac{3^{3x} \times 3^{1 - x}}{3^{2(2 - x)}} = 3^0\) \(3^{3x} \times 3^{1 - x} \div 3^{4x} = 3^0\) \(3^{(3x + 1 - x - 4x)} = 3^0\) \(3^{(1 - 2x)} = 3^0\) since the bases are equal, 1 - 2x = 0 - 2x = -1 x = \(\frac{1}{2}\)

If 2n = y, then, 2\(^{(2 + \frac{n}{3})}\) = 22 x 2\(^\frac{n}{3}\) = 4 x (2n)\(^{\frac{1}{3}}\) But y = 2n, hence 2\(^{(2 + \frac{n}{3})}\) = 4 x y\(^{\frac{1}{3}}\) = 4y\(^\frac{1}{3}\)

tan x = \(\frac{2}{3}\)(given), is illustrated in a right-angled \(\Delta\) thus m2 = 22 + 32 = 4 + 9 = 13 m = \(\sqrt{13}\) Hence, 2sin x = 2 x \(\frac{2}{m}\) 2 x\(\frac{2}{\sqrt{13}}\) = \(\frac{4}{\sqrt{13}}\) = \(\frac{4}{\sqrt{13}} = \frac{\sqrt{13}}{\sqrt{13}}\) = \(\frac{4\sqrt{13}}{13}\)

\(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{2} \times \sqrt{3} + \sqrt{3} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{\sqrt{6} + 3}{3}\) = \(\frac{3 + \sqrt{6}}{3}\) = Hence, (m + n) = 1 + \(\frac{1}{3}\) = 1\(\frac{1}{3}\)

In the diagram |AM| = |MB| - \(\frac{|AB|}{2}\) = \(\frac{16\sqrt{3}}{2}\)cm = 8\(\sqrt{3}\)cm in \(\Delta\) AMO, r2 = |AM|Z + |MO|2 r2 = (8\(\sqrt{3}\))2 + 102 = 64 x 3 + 100 = 192 + 100 = 292 r = \(\sqrt{292}\) 17.088cm 17cm

If log 7.5 = 0.8751 Then 2log75 + log750 = 2(1.8751) + 2.8751 = 3.7502 + 2.8751 = 6.6253

\(= \frac{2\sqrt{2} - \sqrt{3}}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}}\) \(= \frac{2\sqrt{2}(\sqrt{2}) + (2\sqrt{2})(-\sqrt{3})-\sqrt{3}(\sqrt{2})-\sqrt{3}(-\sqrt{3})}{(\sqrt{2})^2-(\sqrt{3})^2}\) \(= \frac{2 \times 2 - 2\sqrt{6} - \sqrt{6} + 3}{2 - 3}\) \(= \frac{4 - 3\sqrt{6} + 3}{-1}\) \(= \frac{7 - 3\sqrt{6}}{-1}\) \(= \frac{7}{-1} - \frac{3\sqrt{6}}{-1}\) \(= -7 + 3\sqrt{6}\) \(= 3\sqrt{6}-7\)

\( = log_2 \frac{8 \times 16}{4}\) \( = log_2 32\) \( = log_2 2^5\) \( = 5log_2 2\) \( = 5 \times 1\) \( = 5 \)

Common ratio r of the G.P is \(r = \frac{T_n + 1}{T_n} = \frac{T_2}{T_1}\) \(r = \frac{\sqrt{10} + 2\sqrt{5}}{\sqrt{10} + \sqrt{5}}\) \(r = \frac{\sqrt{10} + 2\sqrt{5}}{\sqrt{10} + \sqrt{5}} \times \frac{\sqrt{10} - \sqrt{5}}{\sqrt{10} - \sqrt{5}} \) \( = \frac{(\sqrt{10})(\sqrt{10}) + (\sqrt{10})(-\sqrt{5}) + (2\sqrt{5})(\sqrt{10}) + (2\sqrt{5})(-\sqrt{5})}{(\sqrt{10})^2 - (\sqrt{5})^2}\) \(\frac{10 - \sqrt{50} + 2\sqrt{50} - 10}{10 - 5}\) \(\frac{\sqrt{50}}{5}\) \(\frac{\sqrt{25 \times 2}}{5}\) \(\frac{5\sqrt{2}}{5}\) \(\sqrt{2}\)