Question

How do you solve `log_3 (-4x+354)-log_3 (-2x+2)=3`?

Answer 1

I would use the fact that:
`log_aM-log_aN=log_a(M/N)`
You get:
`log_3(-4x+354)-log_3(-2x+2)=3`
`Log_3((-4x+354)/(-2x+2))=3`
Using the definition of logarithm:
`Log_ax=c -> a^c=x`

you get:

`((-4x+354)/(-2x+2))=3^3=27`

and:

`-4x+354=27(-2x+2)`
`-4x+354=-54x+54`
`50x=-300`
`x=-300/50=-6`