Como você condensa # ln2 + 2ln3-ln18 #?

Using the 'laws of logarithms ' :

• logx + logy = logxy

• logx - logy =#log(x/y )#

• # logx^n = n logx #

Applying these to this question :

# ln2 + 2ln3 - ln18 = ln2 + ln3^2 -ln18 = ln2 + ln9 - ln18 #

# = ln((2xx9)/18) = ln(18/18) = ln1 =0#