How do you condense #ln2+2ln3-ln18#?
1 Answer
Jan 15, 2016
0
Explanation:
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#