a diagram shows the (exponential)graph of y=3e^kx. the graph goes through the points (0,3) and (4,18).

what is the value of k?

is it:
a)3/2e

b)1/4 log(smalle)6

c) 1/4 log(smalle)15

d) 1/18 log(smalle)4/3

could you please go over the working because i have no clue what to do...many thanks in advance

Require that y = 3 when x = 0, in order go through the point (0,3). Solve the resulting equation for k, and ALSO make sure the line also goes through the other point.

3 = 3e^(k*0) = 3
This is valid for any k, so it doesn't help at all. Now trying making if go through the other point.
18 = 3e^(4k)
e^(4k) = 6
4k = ln 6
k = (ln6)/4
which is choice b.

Substitute (4,18) into the equation

18 = 3e^(4k)
6 = e^(4k)
take ln of both sides
ln6 = ln(e^(4k))
ln6 = 4k(lne), but lne = 1
ln6 = 4k
k = ln6/4 or the choice of (1/4)lne

(subbing the point (0,3) gains us nothing, since it would be true for any value of k)

To find the value of k in the equation y = 3e^kx, we can use the given points (0,3) and (4,18) to form a system of equations.

First, let's substitute the first point into the equation:

3 = 3e^(k * 0)
3 = 3e^0
3 = 3 * 1
3 = 3

Since this equation holds true, the point (0,3) satisfies the equation.

Next, let's substitute the second point into the equation:

18 = 3e^(k * 4)

Divide both sides of the equation by 3:
18/3 = e^(4k)
6 = e^(4k)

Now, we need to solve for k. To do this, we can take the natural logarithm (ln) of both sides of the equation, since e^x is the inverse of ln(x):

ln(6) = ln(e^(4k))

Using the property of logarithms that ln(a^b) = b * ln(a):

ln(6) = 4k * ln(e)

Since ln(e) = 1, the equation simplifies to:

ln(6) = 4k

Now, divide both sides of the equation by 4:

ln(6)/4 = k

To choose the correct answer from the given options, let's evaluate ln(6)/4:

ln(6) ≈ 1.7918...

Dividing by 4, we get:

ln(6)/4 ≈ 0.4479...

Comparing this approximate value with the options provided:

a) 3/2e ≈ 1.2082...
b) 1/4 log(smalle)6 ≈ 0.4479... (approximate match)
c) 1/4 log(smalle)15 ≈ 0.6607...
d) 1/18 log(smalle)4/3 ≈ 0.0122...

The option that matches the approximate value of ln(6)/4 is b) 1/4 log(smalle)6. Therefore, the value of k is approximately equal to 1/4 log(smalle)6.