jonnburton
Junior Member
- Joined
- Dec 16, 2012
- Messages
- 155
I've managed to find x in terms of t for a differential equation:
\(\displaystyle x = 10 -9e^{\frac{t}{5}}\) and have checked this answer is correct.
But the question (actually this was the first part) asks to find t in terms of x.
I have been trying to do this but cannot get to the answer given in the book.
\(\displaystyle x = 10 -9e^{\frac{t}{5}}\)
\(\displaystyle 10 - x = 9e^{-\frac{t}{5}}\)
\(\displaystyle ln|\frac{10-x}{9}| = -\frac{t}{5}\)
\(\displaystyle t = -5ln|\frac{10-x}{9}|\)
However, the book's answer is \(\displaystyle t = 5ln|\frac{9}{10-x}|\)
In this answer, I don't understand why the negative sign disappears, and how 9 is in the denominator...
\(\displaystyle x = 10 -9e^{\frac{t}{5}}\) and have checked this answer is correct.
But the question (actually this was the first part) asks to find t in terms of x.
I have been trying to do this but cannot get to the answer given in the book.
\(\displaystyle x = 10 -9e^{\frac{t}{5}}\)
\(\displaystyle 10 - x = 9e^{-\frac{t}{5}}\)
\(\displaystyle ln|\frac{10-x}{9}| = -\frac{t}{5}\)
\(\displaystyle t = -5ln|\frac{10-x}{9}|\)
However, the book's answer is \(\displaystyle t = 5ln|\frac{9}{10-x}|\)
In this answer, I don't understand why the negative sign disappears, and how 9 is in the denominator...