She should invest $6491.73.
The equation we use to solve this is in the form
[tex]A=p(1+\frac{r}{n})^{nt}[/tex],
where A is the total amount in the account, p is the principal invested, r is the interest rate as a decimal, n is the number of times per year the interest is compounded, and t is the amount of time.
A in our problem is 14000.
p is unknown.
r is 6% = 6/100 = 0.06.
n is 2, since it is compounded semiannually.
t is 13.
[tex]14000=p(1+\frac{0.06}{2})^{13*2}
\\
\\14000=p(1+0.03)^{26}
\\
\\14000=p(1.03)^{26}
\\
\\\text{Dividing both sides by }1.03^{26}\text{, we have}
\\
\\\frac{14000}{1.03^{26}}=\frac{p(1.03)^{26}}{1.03^{26}}
\\
\\6491.73\approx p[/tex]
