Population growth

[sonrays06]

3. A certain population is growing according to the formula: N = 1410 × 1.03t.

 

[Deleted member 4993]

3. A certain population is growing according to the formula: N = 1410 × 1.03t.

The problem of reference is:

3. A certain population is growing according to the formula: N = 1410 × 1.03t.​

What is the tripling time for this population? Round your answer to one decimal place.​

What is the FIND of the problem - what did the problem ask you to find (calculate)?

 

[HallsofIvy]

I think you mean "1410 x 1.03^t" or \(\displaystyle 1410 \times 1.03^t\)
Do you see that when t= 0 that is 1410?

What t satisfies \(\displaystyle 1410 \times 1.03^t= 3(1410)\)?

 

[sonrays06]

I think you mean "1410 x 1.03^t" or \(\displaystyle 1410 \times 1.03^t\)
Do you see that when t= 0 that is 1410?

What t satisfies \(\displaystyle 1410 \times 1.03^t= 3(1410)\)?

4230

 

[sonrays06]

A certain population is growing according to the formula: N = 1410 × 1.03t.

What is the tripling time for this population? Round your answer to one decimal place.

N=1410 x1.03t

N=1410 x1.03 (3)

N=4356.9

I didn't get the answer right; can someone tell me where I made the mistake?

 

[skeeter]

first off, is your equation [MATH]N = 1410(1.03)^t[/math] ?

if so, what is the population at t = 0?

tripling time is not t = 3, it’s the time required for the initial population to triple in size

 

[sonrays06]

yes, it's 1410(1.03)^t

t= 1410 idk

 

[skeeter]

yes, initial population is 1410

triple that number is 4230

[MATH]4230 = 1410(1.03)^t[/MATH]
can you solve for t ? you’ll need to use logarithms ...

 

[sonrays06]

37.16?

 

[skeeter]

how can you check?

 

[sonrays06]

y=ln(3)ln(1.03)

 

[skeeter]

how about calculating [MATH]1410(1.03)^t[/MATH] using your solution for t ?

shouldn’t you get 4230?