Question 1091449
You invest $ 27,000 for a period of 14 years and rely on an investment value of $ 100,000.
 With which monthly compound annual compound interest rate (12 times per year). You may get this amount at the end of the placement.
:
Using the formula {{{A = Ao(1+(r/n))^(nt)}}} where:
A = final amt
Ao = Initial amt
r = interest rate in decimal form
n = number of times interest is compound per year
t = time in years
in this problem we know:
A = 100000
Ao = 27000
r is unknown
n = 12
t = 14 yrs
:
{{{27000(1+(r/12))^((12*14))) = 100000}}} 
 {{{(1+(r/12))^168 = 100000/27000}}}
 {{{(1+(r/12))^168 = 3.7037}}}
using natural logs, use the log equiv of the exponent
{{{168ln(1+(r/12)) = ln(100000/27000)}}}
find the ln
{{{168ln(1+(r/12)) = 1.309}}}
divide both sides by 168
{{{ln(1+(r/12)) = 1.309/168}}}
{{{ln(1+(r/12)) = .0779365}}}
Find the anti-ln of both sides
1 + {{{r/12}}} = 1.0078241 
subtract 1 from both sides 
 {{{r/12}}} = .0078241
multiply both sides by 12
r =  .0934 or 9.34% interest rate required  
:
A lot of math here; you better check my work. CK