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

f(2x−x2)=−3x2+6x−1

step-by-step explanation:

set up the composite function and evaluate.

f(2x−x2)=−3x2+6x−1

-9

step-by-step explanation: