Question 130665
Steven invests $2000 into a GIC which will double every five years, wheras Dana invests $1000 into a GIC which will triple evry three years. How many years will pass before they have the same amount of money?
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The rule is:

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

We will assume the interest accrues once yearly, so n = 1

and the formula becomes

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

{{{A = P(1+r)^t}}}

So for Steven, P = 2000, so the amount A, in t years, is

{{{A = 2000(1+r)^t}}}

Steven's GIC doubles every five years,
So we substitute t = 5, and A = 4000 (double 2000),
so we can solve for r:

{{{4000 = 2000(1+r)^5}}}

Divide both sides by 2000

{{{2=(1+r)^5}}}

Take the 5th root of both sides:

{{{root(5,2) = root(5,(1+r)^5) }}}

Simplifying the right side:

{{{root(5,2) = 1+r }}}

Solving for r:

{{{root(5,2) - 1 = r}}}

Substituting that into

{{{A = 2000(1+r)^t}}}

{{{A = 2000(1+root(5,2) - 1)^t}}}

or

{{{A = 2000(root(5,2))^t}}}

This is the formula for Steven's amount A in t years.

---------------------------------------------

Now we do exactly the same thing with Dana's
GIC.

{{{A = P(1+r)^t}}}

Now for Dana, P = 1000, so the amount A, in t years, is

{{{A = 1000(1+r)^t}}}

DANA's GIC triples every three years,
So we substitute t = 3, and A = 3000 (triple 1000),
so we can solve for r:

{{{3000 = 1000(1+r)^3}}}

Divide both sides by 1000

{{{3=(1+r)^3}}}

Take the cube root of both sides:

{{{root(3,3) = root(3,(1+r)^3) }}}

Simplifying the right side:

{{{root(3,3) = 1+r }}}

Solving for r:

{{{ root(3,3) - 1 = r}}}

Substituting that into

{{{A = 1000(1+r)^t}}}

{{{A = 1000(1+root(3,3) - 1)^t}}}

or

{{{ A = 1000( root(3,3) )^t}}}

This is the formula for Dana's amount A in t years.

-------------------------------

So to find out when they have an equal amount, we

set their two A's equal, and solve for t:

{{{ A = 2000( root(5,2) )^t }}} and {{{A = 1000( root(3,3) )^t}}}

{{{2000(root(5,2))^t = 1000(root(3,3))^t}}}

Divide both sides by 1000

{{{2(root(5,2))^t = (root(3,3))^t}}}

Change the roots to fractional exponentials of their radicands:

{{{ 2 ( 2^(1/5) )^t }}} = {{{ (3^(1/3))^t  }}}

Multiply inner exponents by outer exponents

{{{2(2^(t/5) ) }}} = {{{3^(t/3)}}}

Write the first {{{2}}} as {{{2^1}}}

{{{2^1(2^(t/5) ) }}} = {{{3^(t/3)}}}

Change the exponent {{{1}}} to {{{5/5}}} so you 
can add exponents on the left:

{{{2^(5/5)(2^(t/5) ) }}} = {{{3^(t/3)}}}

Add exponents:

{{{ 2^(5/5+t/5)  }}} = {{{3^(t/3)}}}

{{{2^((5+t)/5)  }}} = {{{3^(t/3)}}}

{{{2^((5+t)/5 )  }}} = {{{3^(t/3)}}}

Raise both sides to the 15th power

{{{  2^( (5+t)/5 )^15 )    }}} = {{{ ( 3^(t/3) )^15  }}}

Multiply inner exponents by outer exponents:

{{{  2^(  15(5+t) )/5 )}}} = {{{ 3^((15t)/3)}}}

Cancel 5 into 15 on left and 3 into 15 on the right:

{{{2^(3(5+t))}}} = {{{3^(5t)}}}

{{{2^(15+3t)}}} = {{{3^(5t)}}}

Take the log (either log<sub>10</sub> or ln) of both sides

{{{log((2^(15+3t)))}}} = {{{log((3^(5t)))}}}

Use the rule of logs {{{ log(b,x^N) = N*log(b,x)}}}

{{{ (15+3t)log(2) }}} = {{{ (5t)log(3) }}}

Replace {{{log(2)}}} by {{{A}}} and replace {{{log(3)}}} by {{{B}}}

{{{ (15+3t)A }}} = {{{ (5t)B }}}

{{{ A(15+3t)}}} = {{{B(5t)}}}

{{{15A+3At}}} = {{{5Bt)}}}

Isolate the terms in t on the right

{{{15A}}} = {{{5Bt-3At}}}

Factor out t on the right:

{{{15A}}} = {{{t(5B-3A)}}}

Divide both sides by {{{5B-3A}}}

{{{(15A)/(5B-3A)}}} = {{{t}}}

Now replace {{{A}}} by {{{log(2)}}} 
and replace {{{B}}} by {{{log(3)}}}

{{{t}}} = {{{(15log(2))/(5log(3)-3log(2))}}}  

Use calculator to get right side

{{{t}}} = {{{3.045801234}}}

So the answer is just after 3 years.

Edwin</pre></font></b>