Exponential and power functions help

[Intersection]

Hi. Just registered here. I'll go straight to the problems.

1. The question is: From year 2000 to 2010 the amount of units decreased from 22000 to 7000. How many units can we expect at year 2015 if the decrease in % is the same every year?
If Y is the amount of units and X is the amount of years, then where X = 0 Y = 22000 and where X = 10 Y = 7000.

My answer (it's wrong):
f(x) = 22000 * (7000/22000)^x
f(x) = 22000 * 0.3181^x
x * log0.3181 = log 22000
x = log0.3181/log22000 = -0.114553
1 -0.114553 = 0.885447
From here I lost it.

2. A computer can sort out X names every T milliseconds, where T = 1.18 * N^1.18. How many names is sorted in 1 minute?
We know that 1 min = 60000 milliseconds.
60000 = 1.18 * N^1.18
Divide both columns with 1.18
50847.4576 = N^1.18
Here, if it would have been N^2, it would have been easy, but since N^1.18 I have no idea.

 

[Ishuda]

Hi. Just registered here. I'll go straight to the problems.

1. The question is: From year 2000 to 2010 the amount of units decreased from 22000 to 7000. How many units can we expect at year 2015 if the decrease in % is the same every year?
If Y is the amount of units and X is the amount of years, then where X = 0 Y = 22000 and where X = 10 Y = 7000.

My answer (it's wrong):
f(x) = 22000 * (7000/22000)^x
f(x) = 22000 * 0.3181^x
x * log0.3181 = log 22000
x = log0.3181/log22000 = -0.114553
1 -0.114553 = 0.885447
From here I lost it.

2. A computer can sort out X names every T milliseconds, where T = 1.18 * N^1.18. How many names is sorted in 1 minute?
We know that 1 min = 60000 milliseconds.
60000 = 1.18 * N^1.18
Divide both columns with 1.18
50847.4576 = N^1.18
Here, if it would have been N^2, it would have been easy, but since N^1.18 I have no idea.

1. Right idea, just applied wrong. If the percent in decrease is the same each year then the number of items left at the beginning of the next year is the starting amount at the beginning of the year times the factor representing that decrease. So, let a be the (fractional) amount of decrease, i.e. 0.1 or 10%, and f(x) be the amount remaining at the end of x years [that is x = Year - 2000]; Then
f(x) = 22000 (1-a)^x
At the end of 10 years, i.e. beginning in 2010, f(x) is 7000, so
7000 = 22000 (1-a)¹⁰

Can you proceed from there to get a and then calculate the number at the beginning of 2015?


2. This is the same type problem. Just thing of N as a variable which is not necessarily an integer and, like the last problem, note that if
a = b^x
then
b = a^1/x


Just FYI. If you were to approach the problem your way, you could write
f(x) = 22000 * [(7000/22000)^0.1]^x
So that at the end of 10 years we would have
f(10) = 22000 * [(7000/22000)^0.1]¹⁰ = 22000 * (7000/22000)¹ = 22000 * (7000/22000) = 7000
The (7000/22000)^0.1 is the 1-a above and is gen erally easier than logarithms.

 

[Intersection]

1. Right idea, just applied wrong. If the percent in decrease is the same each year then the number of items left at the beginning of the next year is the starting amount at the beginning of the year times the factor representing that decrease. So, let a be the (fractional) amount of decrease, i.e. 0.1 or 10%, and f(x) be the amount remaining at the end of x years [that is x = Year - 2000]; Then
f(x) = 22000 (1-a)^x
At the end of 10 years, i.e. beginning in 2010, f(x) is 7000, so
7000 = 22000 (1-a)¹⁰

Can you proceed from there to get a and then calculate the number at the beginning of 2015?

This is what I got: (7000/22000)^0.1 = 0.892
0.892*22000 = 3960
So the answer is 2015 the amount of units are 4000~ (3960) (which is the correct answer). I don't fully understand it yet. But I think after looking at it some more I will get it. I need to discover it for myself to fully understand it I think.