Grow and doubling time problem. Please help :)

[c47v3770]

Hola!
Hopefully I'm posting in the right section.
Guys, I can't remember how to do this problem.
Would you please help me?

"Oil consumption is growing at a rate of 1.9% per year."
A. What is the doubling time of oil consumption?
B. By what factor will the oil consumption grow in a decade?

Gracias..

 

[skeeter]

let A[sub:2987489g]0[/sub:2987489g] = amount of oil to start with
A = amount of oil at any time t (in years)

A = A[sub:2987489g]0[/sub:2987489g](1.019)[sup:2987489g]t[/sup:2987489g]
edit: should be 1.019 ... time for a new prescription (thanks wjm)

to calculate the doubling time ...

2A[sub:2987489g]0[/sub:2987489g] = A[sub:2987489g]0[/sub:2987489g](1.019)[sup:2987489g]t[/sup:2987489g]
2 = (1.019)[sup:2987489g]t[/sup:2987489g]

now use logarithms to solve for t

to calculate the factor that it will grow in a decade ...

kA[sub:2987489g]0[/sub:2987489g] = A[sub:2987489g]0[/sub:2987489g](1.019)[sup:2987489g]10[/sup:2987489g]

solve for k, the factor.

 

[Denis]

If you're having problems "seeing" why Sir Skeeter's equation (1.19)^t = 2 means the time it takes to "double",
just remember that there is a "1" that doubles to 2, which is not shown: 1(1.19)^t = 2.

 

[c47v3770]

If you're having problems "seeing" why Sir Skeeter's equation (1.19)^t = 2 means the time it takes to "double",
just remember that there is a "1" that doubles to 2, which is not shown: 1(1.19)^t = 2.

Thanks for the reply.

How do I use logarithms to calculate the doubling time? I am a bit confused.

2A0 = A0(1.19)t
2 = (1.19)t

_____________________________
For the grow factor, I got:

kA0 = A0(1.19)10
k= (1.19)10
k= 5.69468

Is that correct?

Thank you..

 

[skeeter]

2 = (1.019)[sup:crje1k6f]t[/sup:crje1k6f] edit for 1.9%

take the log of both sides ...

log(2) = log(1.019)[sup:crje1k6f]t[/sup:crje1k6f]

use the power property of logs on the right side ...

log(2) = t*log(1.019)

log(2)/log(1.019) = t

t = approx 36.8 yrs

If you've never used logs to solve exponential equations, then I recommend you visit the following website ...

http://www.purplemath.com/modules/solvexpo2.htm

 

[Denis]

For the grow factor, I got:
kA0 = A0(1.19)10
k= (1.19)10
k= 5.69468
Is that correct?

Correct: why ask? Do you doubt your calculator :shock:
Start showing powers like this: (1.19)^10

"Growth factor" SIMPLY means the future value of 1
So your 5.69468... simply means that (as example) $1 is worth ~$5.69 after 10 years (if interest = 19% annually).
The formula is simply: (1 + i)^n ; hence (1+ .19)^10 with your problem.

There is no need to confuse the issue with scary-looking stuff like your "kA0 = A0(1.19)10".

Say you're asked: $1500 is deposited in an account earning 7% annually; what's it's value after 13 years?
Step 1: (1 + .07)^13 = ? ; that gives you result for $1
Step 2: multiply above result by 1500 ; that gives you result for $1500

Try it...what do you get?