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| 1. | Find x, if(i) \((\frac{1}{4})^{-4}\times (\frac{1}{4})^{-8}=(\frac{1}{4})^{-4x}\)(ii) \((\frac{-1}{2})^{-19}\div (\frac{-1}{2})^{8}=(\frac{-1}{2})^{-2x+1}\)(iii) \((\frac{3}{2})^{-3}\times (\frac{3}{2})^{5}=(\frac{3}{2})^{-2x+1}\)(iv) \((\frac{2}{5})^{-3}\times (\frac{2}{5})^{15}=(\frac{2}{5})^{2+3x}\)(v) \((\frac{5}{4})^{-x}\div (\frac{5}{4})^{-4}=(\frac{5}{4})^{5}\)(vi) \((\frac{8}{3})^{2x+1}\times (\frac{8}{3})^{5}=(\frac{8}{3})^{x+2}\) | 
| Answer» (i) \((\frac{1}{4})^{-4}\times (\frac{1}{4})^{-8}=(\frac{1}{4})^{-4x}\) \((\frac{1}{4})^{-4}\times (\frac{1}{4})^{-8}=(\frac{1}{4})^{-4x}\) ⇒ \((\frac{1}{4})^{-4-8}\)= \((\frac{1}{4})^{-4x}\)[Using \(a^{n}\times a^{m}=a^{m+n}\)] Equating coefficients when bases are equal. -4-8 = -4x -12 = -4x -12 = -4x x = 3 (ii) \((\frac{-1}{2})^{19}\div (\frac{-1}{2})^{8}=(\frac{-1}{2})^{-2x+1}\) \((\frac{-1}{2})^{19}\div (\frac{-1}{2})^{8}=(\frac{-1}{2})^{-2x+1}\) ⇒ \((\frac{1}{2})^{-19-8}\)= \((\frac{1}{2})^{-2x+1}\)[Using \(a^{n}\div a^{m}=a^{m-n}\)] Equating coefficients when bases are equal. -19-8 = -2x+1 -27 = -2x+1 -27-1 = -2x -28 = -2x x = 14 (iii) \((\frac{3}{2})^{-3}\times (\frac{3}{2})^{5}=(\frac{3}{2})^{2x+1}\) \((\frac{3}{2})^{-3}\times (\frac{3}{2})^{5}=(\frac{3}{2})^{2x+1}\) ⇒ \((\frac{3}{2})^{-3+5}\)= \((\frac{3}{2})^{2x+1}\) [Using \(a^{n}\times a^{m}=a^{m+n}\)] Equating coefficients when bases are equal. -3+5 = 2x+1 2-1 = 2x 1 = 2x x = \(\frac{1}{2}\) (iv) \((\frac{2}{5})^{-3}\times (\frac{2}{5})^{15}=(\frac{2}{5})^{2+3x}\) \((\frac{2}{5})^{-3}\times (\frac{2}{5})^{15}=(\frac{2}{5})^{2+3x}\) ⇒ \((\frac{2}{5})^{-3+15}=(\frac{2}{5})^{2+3x}\)[Using \(a^{n}\times a^{m}=a^{m+n}\)] Equating coefficients when bases are equal. -3+15 = 2+3x 12-2 = 3x 10 = 3x x = \(\frac{10}{3}\) (v) \((\frac{5}{4})^{-x}\div (\frac{5}{4})^{-4}=(\frac{5}{4})^{5}\) \((\frac{5}{4})^{-x}\div (\frac{5}{4})^{-4}=(\frac{5}{4})^{5}\)[Using \(a^{n}\div a^{m}=a^{m-n}\)] Equating coefficients when bases are equal. -x+4 = 5 -x = 5 - 4 -x = 1 x = -1 (vi) \((\frac{8}{3})^{2x+1}\times (\frac{8}{3})^{5}=(\frac{8}{3})^{x+2}\) \((\frac{8}{3})^{2x+1}\times (\frac{8}{3})^{5}=(\frac{8}{3})^{x+2}\)[Using \(a^{n}\times a^{m}=a^{m+n}\)] Equating coefficients when bases are equal. 2x+1+5 = x+2 2x+6 = x+2 2x-x = 2-6 x = -4 | |