hi, i desperately need help solving this question:
Find the values of A and k if the general equation of the graph shown is y=Ae^kx where the graph goes through the points (-1,-4) and (-2,-10)
do i solve it using simultaneous equations? because i have attempted this yet failed :/
Try taking logs of both sides first, after multiplying both sides by -1 (to avoid logs of negative numbers).
ln (-y) = ln (-A) + kx
ln (4) = ln (-A) -k
ln (10) = ln (-A) -2k
-k = ln 10 - ln 4 = 0.91629
k = -0.91629
ln 4 = ln(-A)+0.91629
ln (-A) = 0.47000
A = -1.6000
y = -1.6 e^(-0.91629 x)
To find the values of A and k in the equation y = Ae^kx, you can use the given points (-1,-4) and (-2,-10) to solve a system of equations.
Let's substitute the x and y values of the first point (-1,-4) into the equation:
-4 = Ae^(-k)
Similarly, let's substitute the x and y values of the second point (-2,-10) into the equation:
-10 = Ae^(-2k)
Now we have two equations with two variables (A and k). We can solve this system of equations by isolating A and k individually.
First, let's solve for A:
Divide both sides of the first equation by e^(-k) to isolate A:
-4/e^(-k) = A
Next, let's solve for k:
Divide both sides of the second equation by A to isolate e^(-2k):
-10/A = e^(-2k)
Now, take the natural logarithm (ln) of both sides of the equation to eliminate the exponential term:
ln(-10/A) = -2k
Now, we have the equation ln(-10/A) = -2k, which relates A and k.
To proceed further, we take logs of both sides, after multiplying both sides by -1 (to avoid logs of negative numbers):
ln(-y) = ln(-A) + kx
Now, let's substitute the x and y values of the second point (-2,-10) into this equation:
ln(10) = ln(-A) + 2k
We now have two equations relating A and k:
ln(-10/A) = -2k
ln(10) = ln(-A) + 2k
From the second equation, isolate ln(-A):
ln(-A) = ln(10) - 2k
Now, substitute the value of ln(-A) into the first equation:
ln(-10/A) = -2k
Finally, solve for k:
-k = ln(10/A)
To isolate k, multiply both sides by -1 and divide by 2:
k = -ln(10/A)/2
Now, substitute this value of k back into the second equation (ln(-A) = ln(10) - 2k) to find A:
ln(-A) = ln(10) - 2k
ln(-A) = ln(10) - 2 * (-ln(10/A)/2)
ln(-A) = ln(10) + ln(A)
ln(-A) = ln(10A)
-A = 10A
-1 = 10
This equation is not possible, so there might be an error in the given points or the initial assumptions. Please recheck the problem statement or provide more information.