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.
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
2 answers
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)
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)