Question

How do you solve `log_3 x – 1 = 3`?

Answer 1

Ans: `28`

I think that it is `log_3(x-1)=3`
So you can write:
`3^(log_3(x-1))=3^3`
cancelling the `3` and the `log` on the left you get:
`x-1=27`
`x=28`

The question could also be `log_3(x)-1=3` (with only `x` in the argument although I doubt it) so you get:
`log_3(x)=3+1`
`log_3(x)=4` in the same way as before:
`3^(log_3(x))=3^4`
`x=81`