(x - 2)(x + 4) = (x + 7)(x - 1) High school precalculus. I am learning precalc right now in my math class and I can’t figure out the TWO extraneous solutions to this problem. The topic, the question was under was Logarithmic equations. Originally the problem was “ln(x-2)+ln(x+4)=ln(x+7)+ln(x-1)” I got that down to “ln[(x-2)(x+4)]=ln[(x+7)(x-1)]” then I got that down to “(x-2)(x+4)=(x+7)(x-1)” then that to “x^2+2x-8=x^2+6x-7” my first answer was -1/4=x but then my teacher said I have to solve for extraneous solutions I’m not sure how I find 2 answers to this problem please help

x=-1/4

Step-by-step explanation:

(x-2)(x+4)=(x+7)(x-1)

x^2+2x-8=x^2+6x-7

Subtract x^2 from both sides

2x-8=6x-7

Subtract 6x from both sides

-4x-8=-7

Add 8 to both sides

-4x=1

Divide both sides by -4

x=-1/4

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

step-by-step explanation: hi!