How do you write as a single log for #log_6 60 - log_6 10 #?
1 Answer
Mar 16, 2018
Explanation:
#"using the "color(blue)"laws of logarithms"#
#•color(white)(x)logx-logy=log(x/y)#
#•color(white)(x)log_b b=1#
#rArrlog_6 60-log_6 10=log_6(60/10)=log_6 6=1#
