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

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)

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 ..

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

To find the values of p and q in the equation y = pq^x, given information about the straight line graph of In y against x, we can use the following steps:

(i) Finding the values of p and q:
1. Starting from the given equation y = pq^x, take the natural logarithm of both sides to get ln(y) = ln(pq^x).
2. Apply logarithm properties to simplify the equation: ln(y) = ln(p) + ln(q^x) => ln(y) = ln(p) + x * ln(q).
3. We know that the graph of In y against x is a straight line, so it can be represented by the linear equation: ln(y) = mx + c, where m is the gradient of the line and c is the y-intercept.
4. Comparing the given equation (ln(y) = ln(p) + x * ln(q)) with the linear equation form (ln(y) = mx + c), we can equate the coefficients to find the values of p and q.
a. The coefficient of x in the given equation is ln(q), so q = e^(coefficient of x).
b. The constant term in the given equation is ln(p), so p = e^(constant term).

(ii) Finding the equation relating x and y:
1. Now that we know the values of p and q from step (i), substitute those values into the equation y = pq^x to get the final equation.

Let's calculate the values of p and q and the equation relating x and y:

(i) Calculating p and q:
1. The given gradient of the straight line graph is -3/2.
So, ln(q) = -3/2.
q = e^(-3/2).
Use a calculator to get the decimal value for q.

2. The straight line graph passes through the point (0,2), which we can use to find ln(p).
Let's substitute the coordinates (x, y) = (0, 2) into the equation ln(y) = ln(p) + x * ln(q).
ln(2) = ln(p) + 0 * ln(q) => ln(2) = ln(p).
p = e^(ln(2)).
Use a calculator to get the decimal value for p.

(ii) Calculating the equation relating x and y:
Now we have the values of p and q, we can substitute them into the equation y = pq^x to get the final equation.

The values of p and q (to one decimal place) are determined in step (i), while the equation relating x and y is determined in step (ii).