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if city A's population is 5000 with a yearly growth rate of 4% and city B has a population of 8000 with a yearly growth rate of 2% when will city A surpass city B?

Guest May 2, 2017
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7+0 Answers

 #1
avatar+39 
0

Hold up

ReChan  May 2, 2017
 #2
avatar+39 
0

So equation goes like

5000x+.04

8000x+.02

sub x with 1,2,3,etc

keep going till a passes

ReChan  May 2, 2017
 #3
avatar
0

But what is the answer.

And what is the value of A and B.

Guest May 2, 2017
 #4
avatar+39 
0

I ain't taking away yo education XD Figure it put urself took me long enough to find out

ReChan  May 2, 2017
 #5
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0

I'm doing Java code and I need the answer to know if my code is working and I don't know the answer so I don't know if my code is giving me the right answer.

Guest May 2, 2017
 #6
avatar+39 
+1

Sorry dude got my own stuff to worry bout. Just use a calculator if ur lazy

ReChan  May 2, 2017
 #7
avatar+79835 
+2

 

We want to solve this

 

5000 (1.04)^t   =  8000 (1.02)^t       rearrange as

 

(1.04)^t  / (1/02)^t   =   8000/ 5000

 

(1.04 / 1/02)^t   =  8/5

 

(52/51)^t  =  8/5            take the log of both sides

 

log (52/51)^t   =  log (8/5)     and we can write

 

 t * log (52/51)  =  log (8/5)        divide both sides by log (52/51)

 

t =   log (8/5)  / log (52/51)  =   after  ≈ 24.2 years

 

 

cool cool cool

CPhill  May 2, 2017

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