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, if its linear then
y = Ao ( 1 + .06*t )
if its exponential then
y = Ao ( 1 + .06)^t

i think its exponential
2 = 1 ( 1 + 0.06)^t
now calculate t

Oh, wait. That's right. I apologize. I graphed it wrong.

how will i know if its exponential or not?

here is another example
http://www.physics.uoguelph.ca/tutorials/exp/intro.html

you need a plus somewhere

it should be
next year consumption = last year consumption* ( 1 + 1.9/100)

yes you are write

so we have
x1 = x
x2 = x1 (1.019)
x3 = x2 (1.019) = [x1 (1.019) (1.019) ]

calculate x1,x2 ,x3
and try to find
x2 - x1 = x3 -x2
if its write then only its linear

so did you agree with my math?

right, clearly x2-x1 != x3 - x2

yes i agree
so what you think
its linear or exponential

well... exponential

i think its a little ambiguous however

, you could also have
y = x0 ( 1 + .019 * t ) , no ?

ok how would you word it so that it is linear?

ok , so we wouldnt use percentage for a linear increase

yes

yes

got it,

ok then

ok now doubling time makes sense

doubling time is the time it takes for the initial amount to grow 100%
, leaving you with double

yes
did you get your answer?

yes, the math i didnt have trouble with. just the understanding

i have a much harder problem, differential equation

are you ready?

its a logistic growth problem

ok then
now you can close the qustion
after giving me medal...............lol

you stuck it out , good work

the rest bailed