Question

How to do this?

`log_3 (x-2) = 3-log_3(x+4)`

Answer 1

Rearranging gives:

`log_3(x-2)+log_3(x+4)=3`
which is equivalent to:
`log_3[(x+2)(x+4)]=3`

Since if `N=a^x` then `log_aN=x`
`(x+2)(x+4)=3^3=27`
expanding:
`x^2+6x+8=27`
`x^2+6x-19=0`
`x=-(b+-sqrt(b^2-4ac))/(2a)`

so `x=(-6+-sqrt(36+76))/(2)`
x=2.29 or-8.29