Solve the equation `log_(x^(2)-6x+8)[log_(2x^(2)-2x+8)(x^(2)+5x)]=0`

Options

A. no solution
B. exactly one negative solution
C. at most one negative solution
D. exactly two negative solutions

Answer

B

Explanation

The given equation is meaningful, if
`x^(2)-2x gt0,2x^(2)+2x+3 gt0, 2x^(2)+2x+3 ne1,x^(2)+6x+8 gt0`
and `x^(2)+6x+8 ne 1`
Now,
`x^(2)-2x gt 0 implies x in(-oo,0)cup (2,oo)`
`2x^(2)+2x+3 gt0 impliesx in R " "[:' "Disc"lt 0]`
`2x^(2)+2x+3 ne 1 implies x^(2)+x+1 ne 0 implies x in R`
`x^(2)+6x+8 gt 0 impliesx in(-oo,-4)cup(-2,oo)`
`x^(2)+6x+8 ne 1 implies x^(2)+6x+7 ne 0 implies x^(2)+6x+7 ne 0 implies x ne -3+-sqrt(2)`.
It follows from the above discussion that the given equation is meaningful. if
`x in(-oo,-4)cup(-2,oo)-{-3+sqrt(2),-3-sqrt(2)}`.
For these values of x, we have
`log_(x^(2)+6x+8){log_(2x^(2)+2x+3)(x^(2)-2x)}=0`
`implies log_(2x^(2)+2x+3)(x^(2)-2x)=1`
`implies x^(2)-2x=2x^(2)+2x+3`
`implies x^(2)+4x+3=0 implies(x+3)(x+1)=0 implies x=-1,-3`
But, `x in(-oo,-4)cup(-2,oo)-{-3+sqrt(2),-3-sqrt(2)}`.
Hence, x=-1 is the only solution of the given equation .