How do you simplify #4lne+lne#?

1 Answer
Mar 30, 2016

Answer:

#4ln e + ln e = color(green)(5)#

Explanation:

#ln color(red)(e)#
#color(white)("XXX")=log_color(blue)(e) color(red)(e)#

Remembering that
#color(white)("XXX")log_color(blue)(b) color(red)(a)=c#
means
#color(white)("XXX")color(blue)(b)^c=color(red)(a)#

Therefore
#color(white)("XXX")log_color(blue)(e)color(red)(e)=c#
means
#color(white)("XXX")color(blue)(e)^c=color(red)(e)#

That is
#color(white)("XXX")color(green)(ln_e)=log_e e (= c) = color(green)(1)#

Therefore
#color(white)("XXX")4color(green)(ln_e) + color(green)(ln_e)#

#color(white)("XXX")=4(color(green)(1))+ (color(green)(1))#

#color(white)("XXX")=5#