How do you simplify #[ln(1/4)]/ln2#?

1 Answer
Apr 23, 2016

Answer:

#-2#

Explanation:

#1#. Looking at #ln(1/4)#, express #1/4# in terms of #2#.

#(ln(1/4))/ln2#

#=(ln(2^-2))/ln2#

#2#. Use the natural logarithmic property, #ln_color(purple)e(color(red)m^color(blue)n)=color(blue)n*ln_color(purple)e(color(red)m)#, to simplify the numerator.

#(-2ln2)/ln2#

#3#. #ln2# cancels each other out in the numerator and denominator.

#(-2color(red)cancelcolor(black)(ln2))/color(red)cancelcolor(black)(ln2)#

#=color(green)(|bar(ul(color(white)(a/a)-2color(white)(a/a)|)))#