a c ountrys population in 1991 was 231 million. in 1999 it was 233 million. estimate the population in 2003 using the exponetial growth form

a c ountrys population in 1991 was 231 million. in 1999 it was 233 million. estimate the population in 2003 using the exponetial growth formula. round your answer to the nearest million.

$$\begin{array}{rcl}
(1) \quad p(1991) = 231 &=& p_0 * e^{\lambda*1991}\\
(2) \quad p(1999) = 233 &=& p_0 * e^{\lambda*1999} \\
\hline
\end{array}\\\\
(2):(1)
\begin{array}{rcl}
\frac{233}{231} &=& \frac{ \not{p_0} * e^{\lambda*1999} } {\not{p_0} * e^{\lambda*1991} }\\\\
\frac{233}{231}& = &e^{\lambda*1999-\lambda*1991} = e^{8\lambda}\\\\
\ln{(\frac{233}{231})&=& 8\lambda} \\\\
\lambda &=& \frac{ \ln{(\frac{233}{231} )} } {8} \\\\
\textcolor[rgb]{1,0,0}{ \lambda = 0.00107759288 }
\end{array}\\$$

$$\\p_0 = \dfrac{231}{ e^{\lambda*1991} } = \dfrac{231}{ e^{0.00107759288*1991} } \\\\
\textcolor[rgb]{1,0,0}{p_0= 27.0295384716}$$

exponetial growth formula: $$\boxed{p(year) = 27.0295384716 * e^{ 0.00107759288 * year}}$$

$$\\p(\textcolor[rgb]{1,0,0}{2003}) = 27.0295384716 * e^{ 0.00107759288 * \textcolor[rgb]{1,0,0}{2003} }\\\\
p(2003)= 27.0295384716 * e^{2.15841853962}\\\\
p(2003)= 27.0295384716 * 8.65743543541\\\\
p(2003)= 234.006484167\\\\
\boxed{p(2003)\approx 234 \ Million}$$