We can make use of the general formula for the geometric series to generate the function representing the average annual salary.
an = a0(r)^(n-1)
Or
f(x) = a0(r)^(x - 1)
Plugging in the given values for the year 2005 and 2006 to ge the value of r.
82000 = 70000 (r)^(1-1)
r = 1.1714
Therefore, the function is:
f(x) = 70,000 (1.1714)^(x-1)
Answer:
To work out the multiplier, first add or subtract the percentage from 100, then convert to a decimal. Example: we want to add 20% to £110. To work out the multiplier, add 20 to 100, to get 120, and then change it to a decimal (divide by 100) to get 1.2.
meiabatten191 helped me out on this question too.
Commutative property of addition
This is quite a complex problem. I wrote out a really nice solution but I can't work out how to put it on the website as the app is very poorly made. Still, I'll just have to type it all in...
Okay so you need to use a technique called logarithmic differentiation. It seems quite unnatural to start with but the result is very impressive.
Let y = (x+8)^(3x)
Take the natural log of both sides:
ln(y) = ln((x+8)^(3x))
By laws of logarithms, this can be rearranged:
ln(y) = 3xln(x+8)
Next, differentiate both sides. By implicit differentiation:
d/dx(ln(y)) = 1/y dy/dx
The right hand side is harder to differentiate. Using the substitution u = 3x and v = ln(x+8):
d/dx(3xln(x+8)) = d/dx(uv)
du/dx = 3
Finding dv/dx is harder, and involves the chain rule. Let a = x+ 8:
v = ln(a)
da/dx = 1
dv/da = 1/a
By chain rule:
dv/dx = dv/da * da/dx = 1/a = 1/(x+8)
Finally, use the product rule:
d/dx(uv) = u * dv/dx + v * du/dx = 3x/(x+8) + 3ln(x+8)
This overall produces the equation:
1/y * dy/dx = 3x/(x+8) + 3ln(x+8)
We want to solve for dy/dx, achievable by multiplying both sides by y:
dy/dx = y(3x/(x+8) + 3ln(x+8))
Since we know y = (x+8)^(3x):
dy/dx = ((x+8)^(3x))(3x/(x+8) + 3ln(x+8))
Neatening this up a bit, we factorise out 3/(x+8):
dy/dx = (3(x+8)^(3x-1))(x + (x+8)ln(x+8))
Well wasn't that a marathon? It's a nightmare typing that in, I hope you can follow all the steps.
I hope this helped you :)
ndjdnnx
Step-by-step explanation:
JDUJDNdj
