P grows at the rate of 2% per year, the equation P=1,000,000(1.02)^t, find the value of t for the population is 1500,000
+26400 P grows at the rate of 2% per year, the equation, find the value of t for the population is 1500,000
$$\boxed{P(t)=1000000*(1.02)^t}$$
$$\small{\text{
$
\begin{array}{rcll}
1500000 &=& 1000000*(1.02)^t & \quad | \quad : 1000000\\ \\
1.5 &=& 1.02^t & \\ \\
1.02^t &=& 1.5 & \quad | \quad \ln{()}\\ \\
t*\ln{(1.02)} & = & \ln{(1.5)} \\ \\
t &=& \dfrac{ \ln{(1.5)} }{ \ln{(1.02)} } \\ \\
t &=& \dfrac{ 0.40546510811 } { 0.01980262730 } \\ \\
t &=& 20.4753188576 \\
\end{array}
$
}}$$
The value of t for the population 1500000 is 20.4753188576 years.
+26400 P grows at the rate of 2% per year, the equation, find the value of t for the population is 1500,000
$$\boxed{P(t)=1000000*(1.02)^t}$$
$$\small{\text{
$
\begin{array}{rcll}
1500000 &=& 1000000*(1.02)^t & \quad | \quad : 1000000\\ \\
1.5 &=& 1.02^t & \\ \\
1.02^t &=& 1.5 & \quad | \quad \ln{()}\\ \\
t*\ln{(1.02)} & = & \ln{(1.5)} \\ \\
t &=& \dfrac{ \ln{(1.5)} }{ \ln{(1.02)} } \\ \\
t &=& \dfrac{ 0.40546510811 } { 0.01980262730 } \\ \\
t &=& 20.4753188576 \\
\end{array}
$
}}$$
The value of t for the population 1500000 is 20.4753188576 years.
