Toronto Math Forum
APM3462022S => APM346Lectures & Home Assignments => Chapter 2 => Topic started by: Yifei Hu on February 01, 2022, 10:08:42 AM

When solving this problem, I proceed as follow:
$$\frac{dt}{1}=\frac{dx}{x}=\frac{du}{0}$$
Hence, U does not depend on x and t, integrate on first part of equation:
$$t=ln(x)+C$$
I did not take exponential on both sides to get $e^t=Cx$ but I directly use $C=tln(x)$ and got $U=f(tln(x))$. Can anyone help me identify why this calculation is wrong?

As $x>0$ it is a correct calculation. However $f(xe^{t})$ re,mains valid for $x<0$ while $f(t\ln (x))$ does not.