Question 252185
A GIC pays 6% per annum. How long would it take $3000 to grow to $ 6000?
<pre><font size = 4 color = "indigo"><b>
{{{A = P(1+r/n)^(nt)}}}

We solve for t:

{{{P(1+r/n)^(nt)=A}}}

Take logs of both sides:

{{{log((P(1+r/n)^(nt)))=log((A))}}}

Use a rule of the log of a product:

{{{log((P))+log((1+r/n)^(nt))=log((A))}}}

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

{{{log((1+r/n)^(nt))=log((A))-log((P))}}}

Use a rule of the log of a power:

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


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

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

In your problem, 

P = 3000
A = 6000
r = 6% = .06
n = 1   (per annum)

{{{t = (log((6000))-log((3000)))/(1*log((1+.06/1)))=11.89566105}}}

The 11th year the amount would be 

{{{A = P(1+r/n)^(nt)}}}
{{{A = 3000(1+.06/1)^(1*11)}}}
{{{A = "$5694.89"}}}  [Interest-payers always round down to the lower penny,
                       never up to the higher penny]

The 12th year it would be

{{{A = P(1+r/n)^(nt)}}}
{{{A = 3000(1+.06/1)^(1*12)}}}
{{{A = "$6036.58"}}} 

So the 12th year would be the first time it would
have been at least $6000.

Edwin</pre>