Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

4. write an equation for an exponential growth model with an initial va…

Question

4.
write an equation for an exponential
growth model with an initial value of 5.

Explanation:

Step1: Recall exponential growth formula

The general form of an exponential growth model is \( y = a(1 + r)^t \) or \( y = ae^{kt} \), where \( a \) is the initial value, \( r \) is the growth rate (for the first form), \( k \) is the continuous growth rate (for the second form), and \( t \) is time. Here, we can also use the form \( y = ab^t \) where \( a \) is the initial value and \( b>1 \) for growth.

Step2: Substitute initial value \( a = 5 \)

Using the form \( y = ab^t \) (or \( y = a(1 + r)^t \), or \( y=ae^{kt} \)), substitute \( a = 5 \). A common simple exponential growth model can be written as \( y = 5b^t \) where \( b>1 \), or using the base \( e \) as \( y = 5e^{kt} \) with \( k>0 \), or using the growth rate form \( y = 5(1 + r)^t \) with \( r>0 \). A specific example (taking \( b = 2 \) for simplicity, though any \( b>1 \) works) would be \( y = 5(2)^t \), but the general form with initial value 5 is \( y = 5(1 + r)^t \) (or \( y = 5b^t, b>1 \), or \( y = 5e^{kt}, k>0 \)). The most basic general form for exponential growth with initial value \( a = 5 \) is \( y = 5(1 + r)^t \) where \( r>0 \) is the growth rate, or using the base - multiplier form \( y = 5b^t \) where \( b>1 \). If we take the continuous growth model, it is \( y = 5e^{kt} \) with \( k>0 \). A simple non - specific (since growth rate is not given) exponential growth model with initial value 5 is \( y = 5(1 + r)^t \) (or \( y = 5b^t, b > 1\)) or \( y=5e^{kt},k > 0\). A very common answer (using the base - multiplier form with \( b = e^k\) or just a general base) is \( y = 5(1 + r)^t \) (or \( y = 5b^t\)) with \( a = 5 \). The simplest form when the growth rate is not specified is \( y = 5(1 + r)^t \) (or \( y = 5b^t, b>1\)) or \( y = 5e^{kt},k > 0\). If we assume a general exponential growth model without specifying the growth rate (since it's not given in the problem), we can write \( y = 5(1 + r)^t \) where \( r>0 \) is the growth rate, or \( y = 5b^t \) where \( b>1 \), or \( y = 5e^{kt} \) where \( k>0 \). A standard answer (using the form \( y = ab^t \)) is \( y = 5b^t \) with \( b>1 \), or more commonly, using the growth rate form \( y = 5(1 + r)^t \).

Answer:

A possible exponential growth model equation is \( y = 5(1 + r)^t \) (or \( y = 5b^t, b>1 \), or \( y = 5e^{kt}, k>0 \)). A specific example (taking \( r = 0.1\) for illustration, though any positive \( r \) works) could be \( y = 5(1 + 0.1)^t=5(1.1)^t \), but the general form with initial value 5 is \( y = 5(1 + r)^t \) (where \( r>0 \) is the growth rate) or \( y = 5b^t \) (where \( b > 1\)) or \( y=5e^{kt} \) (where \( k>0\)). The most basic answer (since growth rate is not given) is \( y = 5(1 + r)^t \) (or \( y = 5b^t, b>1 \)) with \( a = 5 \). A common answer is \( y = 5(1 + r)^t \) (or \( y = 5e^{kt} \) or \( y = 5b^t, b>1\)). If we use the base - 2 for simplicity (any \( b>1 \) is fine), \( y = 5(2)^t \) is also a valid exponential growth model with initial value 5.