Question

How do you solve `ln(y-1) - ln(2) = x + ln x`?

Answer 1

Use property of logs ...

Explanation:

`ln[(y-1)/(2xxx)]=x`

Exponentiate ...

`e^{ln[(y-1)/(2xxx)]}=e^x`

Simplify ...

`(y-1)/(2x)=e^x`

`y=(2x)e^x+1`

This function approaches `y=1` as `xrarr-oo`. It never crosses the x-axis.

Hope that helps