How do you solve #120=100(1+(.032/12))^(12t)#?

2 Answers
Nov 14, 2015

Use logarithms.

Explanation:

Divide both sides by 100: #1.2=(1+(.032/12))^(12t)#
Simplify the inside: #1.2=(1.002bar6)^(12t)#
Convert into a logarithm: #log_"1.2"1.002bar6=12t#
Change of base formula: #(log1.002bar6)/log1.2=12t#
Divide both sides by 12: #color(blue)((log1.002bar6)/(12log1.2)=t#

Nov 14, 2015

#t= ln(6/5)/(12ln(1+.032/12))#

Explanation:

We will be using the property of logarithms that
#ln(x^n) = nln(x)#
(this is very useful for solving for variables in exponents)

#120 = 100(1+.032/12)^(12t)#

#=>120/100= 6/5 = (1+.032/12)^(12t)#

#=>ln(6/5) = ln((1+.032/12)^(12t))#

#=> ln(6/5) = 12tln(1+.032/12)# (by the property stated above)

Now, solving for #t# gives

#t = ln(6/5)/(12ln(1+.032/12))#