Evaluate log2 2 8 – log3 3 19

Evaluate log${}_2$ 8 – log${}_3$ $\frac{1}{9}$

Answer Details

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