Rearranging a differential equation

[jonnburton]

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...

 

[HallsofIvy]

Basic law of logarithms: \(\displaystyle \log\left(\frac{a}{b}\right)= log(a)- log(b)= -(log(b)- log(a))= -log\left(\frac{b}{a}\right)\)

 

[srmichael]

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...

Remember that \(\displaystyle \left(\dfrac{a}{b}\right)=\left(\dfrac{b}{a}\right)^{-1}\)

So: \(\displaystyle \left(\dfrac{9}{10-x}\right)=\left(\dfrac{10-x}{9}\right)^{-1}\)

and then the -1 in the exponent comes down in from of the ln and multiplies with the other -1 there to make it positive.

 

[jonnburton]

OK thanks for explaining that, HallsofIvy and Srmichael. I understand now.