log\(_2\) 8 – log\(_3\) \(\frac{1}{9}\)

  = log \(_2\) 2\(^3\) – log\(_3\) 9\(^{-1}\)

  = log\(_2\) 2\(^3\) – log\(_3\) 3\(^{-2}\)

  Based on law of logarithm

  = 3 log\(_2\) 2 – (-2 log\(_3\) 3)

  But log\(_2\) 2 = 1,

  log\(_3\) 3 = 1

  So, = 3 + 2

  = 5