Can someone explain this exponential decay model

[atomic cedar]

@spare laurel

$A = A_{0}e^{kt}$

[spare laurel]

this is what i got

[atomic cedar]

@spare laurel

[spare laurel]

but idk what to do from here

[snow aspen]

you set this up slightly wrong.

[spare laurel]

ye

can u tell me how to set it up

its not in my teacher's notes

[snow aspen]

$0.5A_{0} = A_{0}e^{29k}$

[spare laurel]

OHHHH

ok

so e is constnt

[atomic cedar]

Disorganized

[spare laurel]

$0.5A_{0} = A_{0}e^{29k}$

[atomic cedar]

@spare laurel

[snow aspen]

$0.5 = e^{29k}$

[spare laurel]

$0.5 = e^{29k}$

[atomic cedar]

@spare laurel

[spare laurel]

$ln(0.5) = 29k$

[atomic cedar]

@spare laurel

[snow aspen]

you got it

[spare laurel]

k = ln(0.5) / 29

[atomic cedar]

Disorganized

[snow aspen]

(excuse my typos here)

[spare laurel]

wait i got 0.023

[spare laurel]

so thats k

or the rate at which the substance decas

[snow aspen]

k should be negative

[spare laurel]

now to answer this question

[white kettle]

ln(0.5)/29 ≈ -0,02390162692

[spare laurel]

$0.87 = e^{ \frac{ln(0.5)}{29} \times t}$

[atomic cedar]

@spare laurel

[spare laurel]

$ln(0.87) = \frac{ln(0.5)}{29} \times t$

[atomic cedar]

@spare laurel

[spare laurel]

$t = \frac{29 \times ln(0.87)}{ln(0.5)}$

[atomic cedar]

@spare laurel

[spare laurel]

is this correct @snow aspen

[snow aspen]

looks good

[spare laurel]

k i got ~5.826

[snow aspen]

t will be positive.

[spare laurel]

k is negative if decaying/cooling

k is positive if growing/heating

[snow aspen]

yes, I thought you meant "ok"

before though, you had the wrong value of k

[spare laurel]

actually, i solved my original question before asking here and got negative number

[spare laurel]

but i got confused and thought this was t 😔

[snow aspen]

the insight for how logarithms evaluate is that arguments less than 1 result in negative values, and arguments greater than 1 result in positive values

which is why ln(0.5) -> negative (brain off)

but ln(0.5)/ln(0.67) -> positive (brain off)

[snow aspen]

no big deal

I messed up too

[spare laurel]

_ _

$T = C + (T_{0} - C)e^{kt}$

[atomic cedar]

@spare laurel

[spare laurel]

$56 = 35 + (78 - 35)e^{10k}$

[atomic cedar]

@spare laurel

[spare laurel]

$21 = 43e^{10k}$

[atomic cedar]

@spare laurel

[spare laurel]

$\frac{21}{43} = e^{10k}$

[atomic cedar]

@spare laurel

[spare laurel]

$ln(\frac{21}{43}) = 10k$

[atomic cedar]

@spare laurel

[spare laurel]

$k = \frac{ln( \frac{21}{43})}{10}$