Question 1048483
If Tanisha has $100 to invest at 8% per annum compounded monthly, 
how long will it be before she has $200?
<pre>
{{{A}}}{{{""=""}}}{{{P(1+r/n)^(nt)}}}

{{{200}}}{{{""=""}}}{{{100(1+0.08/12)^(12t)}}}

Divide both sides by 100

{{{2}}}{{{""=""}}}{{{(1+0.08/12)^(12t)}}}

Take natural logs of both sides:

{{{ln(2)}}}{{{""=""}}}{{{ln((1+0.08/12)^(12t))}}}

Use the principle of logs to change an exponent
to a coefficient:

{{{ln(2)}}}{{{""=""}}}{{{12t*ln(1+0.08/12)}}}

{{{ln(2)/(12ln(1+0.08/12))}}}{{{""=""}}}{{{t}}}

{{{8.693188906}}}{{{""=""}}}{{{t}}}

In 8 years and 8 months she will have a little less than $200.
In 8 years and 9 months she will have a little more than $200.
</pre>
If the compounding is continuous, how long will it be?
<pre>
{{{A}}}{{{""=""}}}{{{Pe^(rt)}}}

{{{200}}}{{{""=""}}}{{{100e^(0.08t)}}}

Divide both sides by 100

{{{2}}}{{{""=""}}}{{{e^(0.08t)}}}

Take natural logarithms of both sides

{{{ln(2)}}}{{{""=""}}}{{{ln(e^(0.08t))}}}

{{{ln(2)}}}{{{""=""}}}{{{0.08t}}}

{{{ln(2)/0.08}}}{{{""=""}}}{{{t}}}

{{{8.664339757}}}{{{""=""}}}{{{t}}}

In 8 years and 7 months she will have a little less than $200.
In 8 years and 8 months she will have a little more than $200.

A little bit quicker with continuous compounding but there is
very little difference.

Edwin</pre><b>