How do you find the exponential model #y=ae^(bx)# that goes through the points (0,4) and (5, 1/2)?

Question

Serenity Johnson · Accepted Answer

Solve for #a# using equation 1 and then for #b# using equation 2.

Basically we need to solve for both #a# and #b#. Since #x=0# in the first equation I know that the exponent will be #0# and anything raised to #0# is evaluates to #1# so let's do this first to easily solve for #a#. #4=ae^(b0)# #4=a×1# Having the value of #a# we use the second equation to solve for #b#. #1/2=4e^(b5)# #1/8=e^(b5)# #ln(1/8)=b5# #ln(1/8)/5=b#

Theodore Amidon · Answer

undefined To find the exponential model $ y = ae^{bx} $ that goes through the points (0,4) and (5, 1/2), you need to solve for the constants $ a $ and $ b $ using these points.

1. Plug in the coordinates of the first point (0,4) into the equation:
   $$ y = ae^{bx} $$
   This gives you:
   $$ 4 = ae^{b \times 0} $$
   $$ 4 = a \times e^0 $$
   $$ 4 = a \times 1 $$
   $$ a = 4 $$

2. Now, plug in the coordinates of the second point (5, 1/2) into the equation, using the value of $ a $ you found in the previous step:
   $$ \frac{1}{2} = 4 \times e^{b \times 5} $$
   $$ \frac{1}{8} = e^{5b} $$

3. Take the natural logarithm of both sides to solve for $ b $:
   $$ \ln\left(\frac{1}{8}\right) = \ln(e^{5b}) $$
   $$ \ln\left(\frac{1}{8}\right) = 5b $$
   $$ b = \frac{\ln\left(\frac{1}{8}\right)}{5} $$

Thus, you have found the values of $ a $ and $ b $. Plug these values back into the exponential model equation to get the final equation.