Question 244452
If an investment of $8000 increases by 15 percent at the end of each year, what is the fewest number of years until it doubles in value?
<pre>
If it increases by 15% it becomes 115% of what it was before. To take 115% of 
something is to multiply it by 1.15.  So we want to know how many times
we must multiply $8000 by 1.15 until it just becomes $16000, or first exceeds
it.  Let that number be n.  Then to get the approximate nu,ber of years

{{{8000*1.15^n = 16000}}}

Take the log of both sides:

{{{log(8000*1.15^n) = log(16000)}}}

{{{log(8000)+log(1.15^n) = log(16000)}}}

{{{log(1.15^n) = log(16000)-log(8000)}}}

{{{n*log(1.15) = log(16000)-log(8000)}}}

Divide both sides by {{{log(1.15)}}}

{{{n = (log(16000)-log(8000))/log(1.15)}}}

{{{n = 4.959484455}}}

or in 4 years it would be under $16000 and in 5 years
it would be over $16000.

Edwin</pre>