log2_{2}2 8 – log3_{3}3 19\frac{1}{9}91
= log 2_{2}2 23^{3}3 – log3_{3}3 9−1^{-1}−1
= log2_{2}2 23^{3}3 – log3_{3}3 3−2^{-2}−2
Based on law of logarithm
= 3 log2_{2}2 2 – (-2 log3_{3}3 3)
But log2_{2}2 2 = 1,
log3_{3}3 3 = 1
So, = 3 + 2
= 5
