We know that
a1=1
a2=3
a3=9
a2/a1=3/1----> 3
a3/a2=9/3----> 3
<span>common ration r is equal to 3
number of terms n is 12
The </span><span>Sum of geometric series is given by the formula
</span>Sum=a1*[1-r<span>^n]/[1-r]
</span>Sum=1*[1-3^12]/[1-3]-----> Sum=[1-3^12]/[1-3]----> [3^12-1]/[3-1]
<span>Sum=531440/2-----> 265720
the answer is
265720
</span>
30% of 200=60 200+60= 260
Ty has £260
£260 x 2= £520
200+260+520= £980
Answer £980
I hope this is right :D
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 :)
