Unable to properly manipulate this differential equation

Question

Before the start of a memory test a student has to memorize a set of $60$ facts. According to psychologists the rate at which a person memorizes a set of facts is proportional to the number of facts remaining to be memorized. So if a student memorizes $x$ facts in $t$ minutes, taking $x$ and $t$ as continuous variables, this situation can be modeled by the differential equation $dx/dt = k(60-x)$, where $k$ is a positive constant and $x$ is greater than or equal to $60$ for all $t$ that are greater than or equal to $0$.

a) By solving the differential equation and assuming that initially no facts are memorized, show that $x = 60(1-e^-kt)$

Okay so I don't actually need to do this question, seeing as the equation for the next question is already given here, however I want to be able to recreate what's happening to better understand.

1): $dx/dt = k(60-x)$ (divide both sides by $60-x$)

2): $1/(60-x) dx/dt = k$ (multiply both sides by $dt$)

3): $1/(60-x) dx = k dt$ (integrate both sides)

4): $1/60 ln|-x| = kt + c$ (take exponent from both sides)

5): $1/60-x = e^kt + c$ (multiply both sides by $60$ and rearrange to make $x $ the subject)

$x = 60(-c - e^-kt)$

Where am I going wrong? I assume the arbitrary constant is the $1$ in the given equation, however I get this as a minus integer instead of a positive. Also not sure on how this can be calculated as $1$.

b) Given that the student memorizes fifteen facts in the first twenty minutes, show that $20k = ln3/4$

So going from what was given in a)

$x = 60(1-e^-kt)$, the equation under the given parameters would be

1): $15 = 60(1-e^-20k)$ (take $60$ from both sides)

2): $15/60 = 1-e^-20k$ (take exponential from both sides)

3): $ln15/60 = ln1-(-20k)$ ($ln1 = 0$ and $20k$ becomes positive)

This still leaves $20k = ln1/4$, which again is not the same answer given. If someone could tell me where i'm going wrong so I can properly solve these and any similar questions in the future, I would appreciate it.

Answer 1

For part a, in equation 2, you missed some parentheses. It's not $1/60 -x$ but $1/(60-x)$

Answer 2

There are some mistakes but you were close to the solution

(Line 4 is wrong ... and so is line 5 and 6)

$$dx/dt = k(60-x)$$ $$\frac {dx}{(60-x)}=kdt$$ $$\frac {dx}{(x-60)}=-kdt$$ Integrate both side : $$\ln{(x-60)}=-kt+C$$ $$(x-60)=e^{-kt+C}=Ce^{-kt}$$ $$\boxed{x=Ce^{-kt}+60}$$