x varies partly as y and partly as y^2. When y = 4, x = 52.8, and when y = 5, x = 81. Find x when y = 6.
Let's assume that x varies partly as y and partly as y^2 can be represented by the equation:
x = k * y^a * y^b
where k is a constant and a, b are exponents.
We are given two points that satisfy this equation:
When y = 4, x = 52.8: 52.8 = k * 4^a * 4^b
When y = 5, x = 81: 81 = k * 5^a * 5^b
Now let's solve these two equations to find the values of k, a, and b:
Dividing the second equation by the first equation, we have:
(81/52.8) = (k * 5^a * 5^b) / (k * 4^a * 4^b)
1.5341 = (5^a * 5^b) / (4^a * 4^b)
Since 5 = 4 + 1, we can rewrite the equation as:
1.5341 = (4 + 1)^a * (4 + 1)^b / (4^a * 4^b)
1.5341 = (4 + 1)^(a + b) / 4^(a + b)
1.5341 = 5^(a + b) / 4^(a + b)
Let's define c as a + b:
1.5341 = (5/4)^c
Taking the logarithm of both sides:
log(1.5341) = log((5/4)^c)
log(1.5341) = c * log(5/4)
log(1.5341) / log(5/4) = c
Now, let's plug this value of c back into one of the equations to find the value of k:
52.8 = k * 4^a * 4^b
When y = 4:
k = 52.8 / (4^a * 4^b)
Using c = log(1.5341) / log(5/4), we can rewrite the equation as:
k = 52.8 / (4^a * 4^b) = 52.8 / (4^a * 4^(c - a))
Now let's find the value of b:
k * 5^a * 5^b = 81
Using the previously found value of k, we have:
(52.8 / (4^a * 4^(c - a))) * 5^a * 5^b = 81
52.8 * (5/4)^a * (5/4)^b = 81
52.8 * (5/4)^(a + b) = 81
52.8 * (5/4)^c = 81
Dividing both sides by 52.8, we get:
(5/4)^c = 81 / 52.8
(5/4)^c = 1.5341
Taking the logarithm of both sides:
log((5/4)^c) = log(1.5341)
c * log(5/4) = log(1.5341)
c = log(1.5341) / log(5/4)
Using the given values, we now have the value of c:
c ≈ 1.106
Finally, we can find the value of a:
a = c - b = 1.106 - b
Now we can substitute the values of a, b, and c into k = 52.8 / (4^a * 4^(c - a)) to find k.
Using the values of k, a, b, and c, we can rewrite the equation for x when y = 6 as:
x = k * 6^a * 6^b
Substituting the values, we can calculate x.