y = pq^x
lny = lnp + x lnq
2 = lnp + 0 lnq
p = e^2
1/y y' = lnq
y' = y ln q
-3/2 = 2 lnq
lnq = -3/4
q = e^(-3/4)
y = e^2 e^(-3x/4)
A diagram shows a straight line graph of In y against x,passing through the point (0,2) and having gradient -3/2 .
The variables x and y are connected by the equation y=pq^x ,where p and q are constants
Find
(i)the value of p and of q ,to one decimal place
(ii)the equation relating x and y
3 answers
Can be more specific? still do not understand ...
The answer for part (ii) should be y=7.4(0.2)^2
and the values of q is o.2 ..
The answer for part (ii) should be y=7.4(0.2)^2
and the values of q is o.2 ..
sorry. Guess I should not have given that last line, but the values for p and q were given...
p = e^2 = 7.4
q = e^(-3/2) = 0.2
p = e^2 = 7.4
q = e^(-3/2) = 0.2