Question

How does ln(100-x)^e= (-0.02t + c)^e become 100-x= ce^0.02t??
I got the LHS, but not the RHS

Answer 1

@UnkleRhaukus

Answer 2

\[\ln(100-x)^e= (-0.02t + c)^e\\
e\ln(100-x)= (-0.02t + c)^e\\
\ln(100-x)= (-0.02t + c)^e/e\\
100-x= e^{(-0.02t + c)^e/e}\]
... hmm

Answer 3

I don't get your rhs D:

Answer 4

maybe i made a mistake,

what did you get?

Answer 5

In a chemical reaction a compound X is formed from a compound Y. The masses in grams of X and
Y present at time t seconds after the start of the reaction are x and y respectively. The sum of the two masses is equal to 100 grams throughout the reaction. At any time, the rate of formation of X is proportional to the mass of Y at that time. When t = 0, x = 5 and dx/dt = 1.9.

(i) Show that x satisfies the differential equation: dx/dt = 0.02(100 − x).

(ii) Solve this differential equation, obtaining an expression for x in terms of t. [6]

Answer 6

Was trying to solve ii

Answer 7

Fromm previous ques, Dx/dt =0.02 (100-x)
I got ln (100-x)^e = (-0.02t + c)^e

Answer 8

\[dx/dt = 0.02(100-x)\]

this the variables are separable
\[dx/(100-x) = 0.02dt\]

now integrate
\[\int dx/(100-x) = \int 0.02\,dt\]

Answer 9

Lol I alrd said I got ln(100-x)^e = (-0.02t +c)^e) then what's next

Answer 10

your equation is:
\[\dot x + kx = 100k\]
where k is such that:
\[\dot x = ky\]

Answer 11

@Michele_Laino are we even talking abt the same ques

Answer 12

yes! it is your question

Answer 13

I'm talking abt ii). -.-

Answer 14

ok! I substitute k = 0.02 and I will integrate that equation

Answer 15

How do you freaking integrate it lol. That's my main prob ugh

Answer 16

\[\int dx/(100-x) = \int 0.02\,dt\\
-\ln|100-x|=0.02t+c\]

Answer 17

Uh huh

Answer 18

if you have to integrate this equation:
\[\dot x + P(t)x = Q(t)\]
then your solution x(t) is:
\[\begin{gathered}
x\left( t \right) = \exp \left( { - \int {P(t)dt} } \right)\left( {c + \int {Q(t)\exp \left( {\int {P(t)dt} } \right)dt} } \right) \hfill \\
x(0) = c \hfill \\
\end{gathered} \]

Answer 19

@Michele_Laino Sorry but not sorry. You're totally not helping. I suggest you go help others. Thanks

Answer 20

I'm very sorry, but I suggest you that you have to study about how to integrate some first order ODE's @LXelle

Answer 21

\[\ln|100-x|=-0.02t+c\\
100-x=e^{-0.02t-c}\\
100-x=e^{-0.02t}/e^c\\
100-x=Ce^{-0.02t}\]

Answer 22

first order technique is to general here, it is a variables separable DE

Answer 23

yes! you are right! @UnkleRhaukus nevertheless there are many ways in order to integrate a first order ODE (ordinary differential equation)

Answer 24

keep in mind, this is \(\textit{chemistry}\)

Answer 25

keep in mind that mathematics doesn't distinguish among physics, chemistry etc.. @UnkleRhaukus

Answer 26

@Michele_Laino i expected you to explain and not just throw me some formulas. You weren't even trying to understand my problem

Answer 27

you are right @LXelle nevertheless my formula is a statement of a theorem

Answer 28

Anyways thanks @UnkleRhaukus

Answer 29

please be more polite @LXelle thanks!

Answer 30

Whatever

Answer 31

@LXelle, you could be a little more polite to @Michele_Laino,'s posts Are at least directly relevant to you question.