Question

Solve the equation for W: ln abs(W-300)=(1/25)t+ln(1100).

Answer 1

\[\ln|W-300|=\frac{1}{25}t+\ln1100\]
To rewrite, I'll use the fact that \(\ln e=1\):
\[\ln|W-300|=\frac{1}{25}t\ln e+\ln1100\]
Another property, \(b\ln a=\ln (a^b)\):
\[\ln|W-300|=\ln e^{1/25~t}+\ln1100\]
Another property, \(\ln a+\ln b=\ln (ab)\):
\[\ln|W-300|=\ln 1100e^{1/25~t}\]
Now you can take the exponential of both sides:
\[e^{\ln|W-300|}=e^{\ln 1100e^{1/25~t}}\]
which reduces to the simpler equation,
\[|W-300|=1100e^{1/25~t}\]

Answer 2

Thank you.

Answer 3

yw