Question

(2(x+1)(x+2)(x+3))^x² = (3(x+4)(x+2)(x+3))^x² find all values of x

Answer 1

x = -10, -3, -2, 0

Answer 2

We are given

$$\Bigl[ 2(x+1)(x+2)(x+3) \Bigr]^{x^2} = \Bigl[ 3(x+4)(x+2)(x+3) \Bigr]^{x^2}.$$

Step 1. Notice that the same exponent $x^2$ appears on both sides.

Case 1: $x^2 = 0$
This gives

$$x = 0.$$

When $x = 0$, the left‐side becomes $[2(1)(2)(3)]^0 = 12^0 = 1$ and the right side becomes $[3(4)(2)(3)]^0 = 72^0 = 1$ so the equality holds.

Case 2: $x^2 \neq 0$
Since the exponent is nonzero, we may take logarithms (or notice that if $a^{x^2} = b^{x^2}$ with $x^2 \neq 0$ then $a = b$, provided $a$ and $b$ do not vanish in an ambiguous way). Thus, we set

$$2(x+1)(x+2)(x+3) = 3(x+4)(x+2)(x+3).$$

If $(x+2)(x+3) \neq 0$, divide both sides by $(x+2)(x+3)$:

$$2(x+1) = 3(x+4).$$

Now, solve for $x$:

$$2x + 2 = 3x + 12 \quad \Longrightarrow \quad 2 = x + 12 \quad \Longrightarrow \quad x = -10.$$

Case 2a: When one of the factors $(x+2)$ or $(x+3)$ is zero.

1. If $x+2 = 0$ then $x = -2$.
2. If $x+3 = 0$ then $x = -3$.

For these values, notice:

• At $x=-2$:
    Left-hand side:   $2(-2+1)(0)(-2+3) = 2(-1)(0)(1) = 0.$   Right-hand side:   $3(-2+4)(0)(-2+3) = 3(2)(0)(1) = 0.$   Since $x^2=4>0$, we have $0^4=0$ on both sides.
• At $x=-3$:
    A similar calculation shows both sides become $0^9 = 0$.

Thus, $x=-2$ and $x=-3$ are also solutions.

Thus, the complete set of solutions is $\boxed{-10, -3, -2, 0}$.