Come si condensano # ln2 + 2ln3-ln18 #?
Risposta:
0
Spiegazione:
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#