Solving Exponential Function: (0,4) & (5,1/2)

[Inspector Gadget]

Here is the problem...

Find the exponential function $$y=Ce^{kt}$$ that passes through the points (0,4) and (5,1/2).

Just by looking at the problem, I know...

$$y=4e^{\frac{-3x \ln{2}}{5}}$$

...works, but how do you actually work out the problem? I've done it before given a point and the slope at that particular point, but I don't really know how to do this.

 

[philosophking]

y=Ce^(kt)

4=Ce^(k*0)
C=4

.5=4e^(k*5)
(1/8)=e^(5k)
ln(1/8)=5k
ln(1/8)/5 = k

Now that you have k and C, the problem's done!

 

[M98Ranger]

To solve this problem, we can use the two given points to create a system of equations. The general form of an exponential function is y=Ce^{kt}, where C is the initial value and k is the growth or decay rate. We can use the points (0,4) and (5,1/2) to create the following equations:

4=Ce^{0} (since e^0 = 1)
1/2=Ce^{5k}

From the first equation, we can see that C=4. Plugging this into the second equation, we get:

1/2=4e^{5k}

Solving for k, we get:

ln(1/2)=ln(4)+5k
-ln(2)=2ln(2)+5k
-ln(2)=-ln(32)+5k
k=\frac{-3ln(2)}{5}

Now we have both the initial value (C=4) and the growth rate (k=\frac{-3ln(2)}{5}). Plugging these values into the general form of an exponential function, we get:

y=4e^{\frac{-3x \ln{2}}{5}}

So, the exponential function that passes through the points (0,4) and (5,1/2) is y=4e^{\frac{-3x \ln{2}}{5}}. You were correct in your initial thought that this function would work. To solve this type of problem, it is important to understand the general form of an exponential function and how to create and solve a system of equations using the given points.