Solve the Log Inequality: 2log₃x < log₃(x²+2x-12)

Let's solve the inequality $2\log_3x < \log_3(x^2+2x-12)$.

• Step 1: Apply the Power Property of Logarithms

The expression $2\log_3x$ can be rewritten as $\log_3(x^2)$ using the power property, which states $a\log_b(x) = \log_b(x^a)$.

Thus, the inequality transforms to:

$\log_3(x^2) < \log_3(x^2 + 2x - 12)$

• Step 2: Remove the Logarithm by Ensuring Both Sides are Positive

Since $\log_3(M) < \log_3(N)$ implies $M < N$ when $M > 0$ and $N > 0$, the inequality becomes:

$x^2 < x^2 + 2x - 12$

Simplifying:

$0 < 2x - 12$

Add 12 to both sides:

$12 < 2x$

Divide both sides by 2:

$6 < x$

• Step 3: Consider the Domain Restrictions of the Logarithmic Terms

For both sides of the logarithmic inequality to be defined, we need to ensure:

• $x > 0$
• Expression inside the right logarithm is positive: $x^2 + 2x - 12 > 0$

Solving $x^2 + 2x - 12 > 0$ involves factorization:

$(x + 4)(x - 3) > 0$

This quadratic inequality gives critical points at $x = -4$ and $x = 3$. Testing intervals around these points, the inequality holds when $x < -4$ or $x > 3$. Considering the logarithmic condition $x > 0$, we narrow it to $x > 3$.

• Step 4: Combine All Results

The combined condition from steps 2 and 3 yield:

$6 < x$

Therefore, the solution to the inequality is $\boxed{6 < x}$.