Find the exponential model $y = a e ^ { b x }$ that fits the points shown in the table. $$\begin{array} { | l | l | l | }
\hline x & 0 & 4 \
\hline y & 5 & 1 \
\hline
\end{array}$$
A) $y = 5 e ^ { - 0.4024 x }$
B) $y = 5 e ^ { 0.5024 x }$
C) $y = - \frac { 1 } { 5 } e ^ { 0.4024 x }$
D) $y = \frac { 1 } { 5 } e ^ { - 0.6024 x }$
E) $y = - \frac { 1 } { 5 } e ^ { - 0.4024 x }$

Question

Find the exponential model $y = a e ^ { b x }$ that fits the points shown in the table. $$\begin{array} { | l | l | l | } 
\hline x & 0 & 4 \
\hline y & 5 & 1 \
\hline
\end{array}$$
A) $y = 5 e ^ { - 0.4024 x }$
B) $y = 5 e ^ { 0.5024 x }$
C) $y = - \frac { 1 } { 5 } e ^ { 0.4024 x }$
D) $y = \frac { 1 } { 5 } e ^ { - 0.6024 x }$
E) $y = - \frac { 1 } { 5 } e ^ { - 0.4024 x }$

Quizplus · Accepted Answer

The exponential model is $y = a e^{bx}$. Given $y = 5$ when $x = 0$, we can substitute these values into the equation to find $a$, resulting in $5 = a e^{0}$, which simplifies to $a = 5$. For $x = 4$ and $y = 1$, substituting these values into the equation $1 = 5e^{4b}$ allows us to solve for $b$. Dividing both sides by 5 gives $e^{4b} = \frac{1}{5}$. Taking the natural logarithm of both sides, we get $4b = \ln(\frac{1}{5})$, which simplifies to $b = \frac{\ln(\frac{1}{5})}{4}$. This calculation results in $b \approx -0.4024$, making the correct model $y = 5 e^{-0.4024x}$.

Find the Exponential Model y=aebxy = a e ^ { b x }y=aebx

Find the Exponential Model $y = a e ^ { b x }$