Question

How do you solve `ln(2x -1) = 0`?

Answer 1

`x=1`

Explanation:

`ln(2x-1)=0`
We know that, `AAainR^(+)-{1},x inR, and, y inR^(+)color(red)(log_ay=XhArry=a^X)`
So,
`ln(2x-1)=0hArr(2x-1)=e^0=1`
`rArr2x-1=1rArr2x=2rArrx=1`
Checking: Put x=1
`LHS=ln(2(1)-1)=ln(2-1)=ln1=0=RHS`