Question

What is x if `log_3 (2x-1) = 2 + log_3 (x-4)`?

Answer 1

`x = 5`

Explanation:

We will use the following:

• `log_a(b) - log_a(c) = log_a(b/c)`
• `a^(log_a(b)) = b`

---

`log_3(2x-1) = 2 + log_3(x-4)`

`=> log_3(2x-1) - log_3(x-4) = 2`

`=>log_3((2x-1)/(x-4)) = 2`

`=> 3^(log_3((2x-1)/(x-4))) = 3^2`

`=> (2x-1)/(x-4) = 9`

`=> 2x - 1 = 9x - 36`

`=> -7x = -35`

`=> x = 5`

Answer 2

I found: `x=5`

Explanation:

We can start writing it as:
`log_3(2x-1)-log_3(x-4)=2`
use the property of the logs: `logx-logy=log(x/y)` and write:
`log_3((2x-1)/(x-4))=2`
use the definition of log:
`log_bx=a->x=b^a`
to get:
`(2x-1)/(x-4)=3^2` rearranging:
`2x-1=9(x-4)`
`2x-9x=-36+1`
`7x=35`
`x=35/7=5`