Question 543333
You're looking for the compounding formula:
{{{A=p*(1+(i/m))^(m*n)}}}
where
"A" = amount you will ultimately have in the account (A=1200+200=1400)
"P" = amount you started with, aka principal (p=1200)
"i" = interest rate per year (i=9%)
"m" = number of compounding periods per year (Quarterly means m=4)
"n" = number of years (n=??? - this is what you're solving for)
Plug into the formula and solve for N.
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{{{A=p*(1+(i/m))^(m*n)}}}
{{{1400=1200*(1+(.09/4))^(4*n)}}}
{{{1400=1200*(1.0225)^(4*n)}}}
{{{1400/1200=(1.0225)^(4*n)}}}
{{{7/6=(1.0225)^(4*n)}}}
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Here's where most people give up. They don't remember how to bring the variable in the exponent down to work with. The solution -> natural log function!
{{{ln(n^x)=x*ln(n)}}}
Example: Take {{{5^x}}}
{{{ln(5^x)=x*ln(5)}}}
and {{{ln(5)=1.609}}}
So {{{x*ln(5)=1.609x}}}
It allows you to bring the exponent down so it's workable.
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So what we need to do now is apply the ln function to both sides of the equation. And remember ln(some number) is a constant. You can calculate it or divide by it.
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{{{ln(7/6)=ln((1.0225)^(4*n))}}}
{{{ln(7/6)=(4*n)*ln(1.0225)}}}
{{{ln(7/6)/ln(1.0225)=4*n}}}
{{{n=(1/4)(ln(7/6)/ln(1.0225))}}}
Time to bust out the calculator
{{{n=(1/4)(ln(7/6)/ln(1.0225))=(1/4)(6.9279)=1.732}}}
So it takes approx 1.732 years (20.78 months) to earn $200 of interest.