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
