Question

Solve the equation algebraically.
ln (y-1) - ln 2 = x + ln x

Answer 1

for x or y?

Answer 2

for x, sorry

Answer 3

x=ln((y-1)/2x)

Answer 4

\[\ln{(y-1)-\ln{2}}=x+\ln{x}\]
\[\ln{\frac{y-1}{2}}=x+\ln{x}\]
\[\frac{y-1}{2}=\exp{(x+\ln{x})}=x\cdot e^x\]
\[y=2x\cdot e^x+1\]