Log2(x-3)+log2x-log2(x+2)=2

We are given the equation:

log₂(x-3) + log₂x - log₂(x+2) = 2

When is it defined:

in this equation, log₂(x-3) and log₂(x+2) can only be defined when

x-3 >0 and x+2 > 0

solving for the values of x, we get:

x > 3    and   x > -2

which basically means x > 3

Because we are looking for an inequality which is true for both x>-2 and x>3

Hence, x will have a value greater than 3

Solving for x:

using the product rule [logₐb + logₐc = logₐ(bc)]

log₂[(x-3)(x)] - log₂(x+2) = 2  

using the quotient rule [logₐb - logₐc = logₐ(b/c)]

log₂[(x-3)(x) / (x+2)] = 2

from the property [ aˣ = b  ⇒  logₐb = x]

(x² - 3x) / (x+2) = 2²

x² - 3x = 4x + 8

x² - 7x - 8 = 0

x² + x - 8x - 8 = 0

x(x+1) - 8(x+1) = 0

(x-8)(x+1) = 0

(x-8) = 0    OR    (x+1) = 0

x = 8      OR      x = -1

We know that the equation is defined only for x > 3

We can see that x = 8 satisfies that inequality

Therefore, x = 8

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?