Question 878997
<pre>
It's a geometric series with a<sub>1</sub> = 1 and r = 3.

The sum of a geometric series is given by the formula:

{{{S[n]}}}{{{""=""}}}{{{a[1](r^n-1)/(r-1)}}}

We set that greater than 7000

{{{a[1](r^n-1)/(r-1)}}}{{{"">""}}}{{{7000}}}

{{{1(3^n-1)/(3-1)}}}{{{"">""}}}{{{7000}}}

{{{(3^n-1)/2}}}{{{"">""}}}{{{7000}}}

Multiply both sides by 2 to clear the fraction:

{{{3^n-1}}}{{{"">""}}}{{{14000}}}

Add 1 to both sides

{{{3^n}}}{{{"">""}}}{{{14001}}}

Take the ln or log of both sides:

{{{log((3^n)))}}}{{{"">""}}}{{{log((14001))}}}

Use the property of logs that says that the log of 
an exponential is the product of the exponent and 
the log of the base.

{{{n*log((3))}}}{{{"">""}}}{{{log((14001))}}}

Divide both sides by log(3)

{{{n}}}{{{"">""}}}{{{log((14001))/log((3))}}}

{{{n}}}{{{"">""}}}{{{8.68994834}}}

The least value of n greater than that is when n=9.

Edwin</pre>