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



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