Question #32dad

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Answer 1 · 2018-01-15T17:10:42

#P(1+x%)^n#

#n~~ 7.761....# years

Rounded to 8 years

#P(1+x%)^n# for yearly calculation cycle#" "...Equation(1)#

Note that for monthly calculation cycle you would have:
#P(1+(x%)/12)^(12n)#

Using #Equation(1)# we have:

#P(1+x%)^n color(white)("d") -> color(white)("d")$2500(1+3.2/100)^n=($2500+$716.46)#

#(103.2/100)^n= (cancel($)3216.46)/(cancel($)2500)# Notice the units cancel out.

Take logs of both sides

#nxx ln(103.3/100)=ln( 3216.46/2500)#

#n=ln( 3216.46/2500)-: ln(103.3/100)#

#n~~ 7.761....#