Question

describe a population whose amount a doubles every year using log

Explanation:

Step1: Define the exponential growth model

Let the initial population be \( A_0 \) and the population after \( t \) years be \( A(t) \). For a population that doubles every year, the exponential growth formula is \( A(t)=A_0\times2^t \), since the growth factor is 2 (doubling) per year.

Step2: Apply logarithms to solve for time or analyze

If we want to find the time \( t \) when the population reaches a certain amount \( A \), we can take the logarithm of both sides. Let's use the natural logarithm (ln) or common logarithm (log base 10). Using natural logarithm:
\( \ln(A(t))=\ln(A_0\times2^t) \)
By the logarithm property \( \ln(ab)=\ln(a)+\ln(b) \), we get:
\( \ln(A(t))=\ln(A_0)+t\ln(2) \)
We can solve for \( t \):
\( t = \frac{\ln(A(t))-\ln(A_0)}{\ln(2)} \)
This formula allows us to find the time it takes for the population to grow from \( A_0 \) to \( A(t) \) using logarithmic operations. For example, if the initial population \( A_0 = 100 \) and we want to find when it reaches \( A(t)=800 \), we substitute into the formula:
\( t=\frac{\ln(800)-\ln(100)}{\ln(2)}=\frac{\ln(\frac{800}{100})}{\ln(2)}=\frac{\ln(8)}{\ln(2)} \)
Since \( 8 = 2^3 \), \( \ln(8)=3\ln(2) \), so \( t = \frac{3\ln(2)}{\ln(2)} = 3 \) years. This shows how logarithms help in analyzing the exponential growth (doubling population) by converting the exponential equation into a linear form in terms of \( t \) when we take the log.

Answer:

The population follows an exponential growth model \( A(t)=A_0\times2^t \), and logarithms (e.g., natural log or common log) are used to solve for time \( t \) (or analyze the growth) via \( t=\frac{\ln(A(t)) - \ln(A_0)}{\ln(2)} \) (for natural log) to understand when the population reaches a certain size, leveraging logarithmic properties to transform the exponential relationship into a solvable linear - like form for \( t \).