Question 381329
3. How long, to the nearest tenth of a year, will it take $12,500 to grow to $20,000 at 6.5% annual interest compounded quarterly? (Use the formula for compound interest with n compoundings per year to solve for t.)
<pre>
{{{A}}}{{{""=""}}}{{{P(1+r/n)^(nt)}}}

A = final amount = 20000
P = beginning amount = 12500
r = annual interest rate expressed as a decimal = .065
n = number of times a year the interest is compounded = 4 (quarterly)
t = number of years it takes P to grow to A = what we want to find

{{{A}}}{{{""=""}}}{{{P(1+r/n)^(n*t)}}}

Take the log of both sides:

{{{log((A))}}}{{{""=""}}}{{{log((P*(1+r/n)^(n*t)))}}}

Use this rule of logs:  {{{log((X*Y))=log((X))+log((Y))}}}

{{{log((A))}}}{{{""=""}}}{{{log((P))+log( (1+r/n)^(n*t))   }}}

Subtract {{{log((P))}}} from both sides:

{{{log((A))-log((P))}}}{{{""=""}}}{{{n*t*log( (1+r/n))   }}}

Divide both sides by {{{n*log((1+r/n))}}}

{{{(log((A))-log((P)))/(n*t*log( (1+r/n)))}}}{{{""=""}}}{{{(n*t*log( (1+r/n)))/(n*log( (1+r/n)))}}}

Cancel:

{{{(log((A))-log((P)))/(n*t*log( (1+r/n)))}}}{{{""=""}}}{{{(cross(n)*t*cross(log( (1+r/n))))/(cross(n)*cross(log( (1+r/n))))}}}

{{{(log((A))-log((P)))/(n*log( (1+r/n)))}}}{{{""=""}}}{{{t}}}

{{{t}}}{{{""=""}}}{{{(log((A))-log((P)))/(n*log( (1+r/n)))}}}

Now substitute

A = final amount = 20000
P = beginning amount = 12500
r = annual interest rate expressed as a decimal = .065
n = number of times a year the interest is compounded = 4 (quarterly)

{{{t}}}{{{""=""}}}{{{(log((20000))-log((12500)))/(4*log( (1+.065/4)))}}}

{{{t}}}{{{""=""}}}{{{(log((20000))-log((12500)))/(4*log( (1+.065/4)))}}}

{{{t}}}{{{""=""}}}{{{7.289417683}}}

That's over 7 years.  To find out how many months, multiply the
fractional part .289417683 by 12, getting 3.473012194

So 7 years 3 months the amount will be less than $20000 and 7 years
4 months it will be more than $20000.

Edwin</pre>