How do you condense #log_4z-log_4y#?

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Answer 1 · 2018-04-07T13:46:07

#color(blue)(log_4(z/y)#

By the laws of logarithms:

#log_c(a/b)=log_c(a)-log_c(b)#

#:.#

#log_4z-log_4y=log_4(z/y)#

#color(blue)(log_4(z/y)#

Answer 2 · 2018-04-07T13:49:54

#log_4(z/y)#

#"using the "color(blue)"law of logarithms"#

#•color(white)(x)loga-logb=log(a/b)#

#rArrlog_4z-log_4y=log_4(z/y)#