Question

How do you solve `log (x+3)=1-logx`?

Answer 1

`=>x=-5 and x=2`

Explanation:

`log(x+3)+logx =1`
`=>log((x+3)xxx) =log10`
`=>x^2+3x=10`
`=>x^2+3x-10=0`
`=>x^2+5x-2x-10=0`
`=>x(x+5)-2(x+5)=0`
`=>(x+5)(x-2)=0`
`=>x=-5 and x=2`