How do you simplify #\log _{3}8+\log _{3}7+\log _{3}11#?

2 Answers
Apr 1, 2018

The expression is equivalent to #log_3 616# which is about #5.84669...#

Explanation:

You can simplify this because it's just a bunch of numbers, so you can condense it using this logarithm rule:

#log_color(green)acolor(red)x+log_color(green)acolor(blue)y=log_color(green)a(color(red)xcolor(blue)y)#

Here's our expression:

#color(white)=log_3 8+log_3 7+log_3 11#

#=log_color(magenta)3 color(red)8+log_color(magenta)3 color(blue)7+log_color(magenta)3 color(green)11#

#=log_color(magenta)3 (color(red)8*color(blue)7*color(green)11)#

#=log_color(magenta)3 (color(purple)56*color(green)11)#

#=log_color(magenta)3 (color(brown)616)#

#~~5.84669...#

Apr 1, 2018

#log_3 616#

Explanation:

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

#•color(white)(x)logx+logy=log(xy);x,y>0#

#"this can be extended to more than 2 terms"#

#rArrlog_3 8+log_3 7+log_3 11#

#=log_3(8xx7xx11)=log_3 616#