`log_5(2x-7)=log_5(3x-9)` Solve the equation. Check for extraneous solutions.

`log_5(2x-7)=log_5(3x-9)`

Using one to one property of logarithms,

`2x-7=3x-9`

`=>2x-3x=-9+7`

`=>-x=-2`

`=>x=2`

Let's plug back the solution in the equation,

`log_5(2*2-7)=log_5(3*2-9)`

`log_5(-3)=log_5(-3)`

However logarithm of negative number is undefined,

So the solution is extraneous.

`log_5(2x-7)=log_5(3x-9)`

Using one to one property of logarithms,

`2x-7=3x-9`

`=>2x-3x=-9+7`

`=>-x=-2`

`=>x=2`

Let's plug back the solution in the equation,

`log_5(2*2-7)=log_5(3*2-9)`

`log_5(-3)=log_5(-3)`

However logarithm of negative number is undefined,

So the solution is extraneous.