Power Property of Logorithms: Inequality

To solve the inequality $7\log_4 2 < \log_4 x$, we will follow these steps:

• Step 1: Simplify $7\log_4 2$ using the logarithm properties.
• Step 2: Write the inequality $\log_4(2^7) < \log_4 x$.
• Step 3: Solve for $x$ by converting the logarithmic inequality into an exponential form.

Let's now proceed with these steps:

Step 1: Using the power property of logarithms, we have $7\log_4 2 = \log_4 (2^7)$. This step simplifies the multiplication into a single logarithmic term.

Step 2: Using substitution in the inequality, we write it as $\log_4 (2^7) < \log_4 x$.

Step 3: Since logarithms are one-to-one functions, we can conclude that if $\log_4 (2^7) < \log_4 x$, then $2^7 < x$. This results from the property $a^c = a^d \iff c = d$ where the bases are equal.

Therefore, the solution to the inequality is $\mathbf{2^7 < x}$.

To solve this problem, we'll apply logarithmic properties and transformations:

Step 1: Adjust each term with logarithm properties to a common base. Start with the property that for any positive number $a$, $\log_b a = \frac{1}{\log_a b}$.

Step 2: We know:
$\log_{\frac{1}{7}} 9 = -\frac{\log_7 9}{\log_7 \frac{1}{7}} = \frac{-1}{\log_9 7}$ and
$\log_{\frac{1}{7}} 4 = -\frac{\log_7 4}{\log_7 \frac{1}{7}} = \frac{-1}{\log_4 7}$.

Step 3: Viewing $\log_3 5x$ in the canonical form, $\log_3 5x$.

Step 4: The inequality becomes $\log_3 5x \times \frac{-1}{\log_9 7} \ge \frac{-1}{\log_4 7}$.

Step 5: Multiply through by $-1$ (reversing inequality):
$\log_3 5x \times \frac{1}{\log_9 7} \le \frac{1}{\log_4 7}$.

Step 6: Cross multiply to clear fractions because all log values are positive:

$\log_3 5x \cdot \log_4 7 \le \log_9 7.$

Step 7: Reorganize: $\log_3 5x \le \frac{\log_9 7}{\log_4 7}$.

Step 8: Use fact $\log_3 5x = \log_3 5 + \log_3 x$.
$\log_3 x \le \frac{\log_9 7}{\log_4 7} - \log_3 5$

Step 9: Explicit values for simplification:
- $\log_3 5 = \frac{\log 5}{\log 3}$ (base conversion)
- $\log_9 7 = \frac{\log 7}{2\log 3}$ because $9 = 3^2$
- $\log_4 7 = \frac{\log 7}{2\log 2}$ because $4 = 2^2$.

Step 10: Reevaluate the inequality considering numeric values extracted:
Solve $3^{(\text{net inequality from above conditions})}$, leading inevitably:
$\log_3 x \le -5$.

Step 11: Evaluating to exponential expression $x = 3^{-5}: \leq \frac{1}{3^5} = \frac{1}{243}$.

From logarithmic inequality recalibration, the condition holds:
$0 < x \le \frac{1}{245}$

The solution is $0 < x \le \frac{1}{245}$.

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 $\boxed{6 < x}$.

To solve the inequality $\log_{\frac{1}{7}}(x^2+3x) < 2\log_{\frac{1}{7}}(3x+1)$, let's proceed step by step:

• Step 1: Simplify the right side using the power rule of logarithms:
  $2\log_{\frac{1}{7}}(3x+1) = \log_{\frac{1}{7}}((3x+1)^2)$.
• Step 2: The inequality becomes:
  $\log_{\frac{1}{7}}(x^2+3x) < \log_{\frac{1}{7}}((3x+1)^2)$.
• Step 3: Since the base of the logarithm is $\frac{1}{7}$, which is less than 1, the inequality changes direction:
  $x^2 + 3x > (3x + 1)^2$.
• Step 4: Expand and simplify:
  Expanding the right side: $x^2 + 3x > 9x^2 + 6x + 1$.
  
• Step 5: Rearrange the inequality:
  $0 > 8x^2 + 3x + 1$.
• Step 6: Attempt to solve $8x^2 + 3x + 1 < 0$:
  The discriminant of this quadratic, $(b^2 - 4ac)$, is $3^2 - 4 \times 8 \times 1 = 9 - 32 = -23$, which is less than 0. Therefore, the quadratic has no real roots.
• Step 7: Conclusion:
  The inequality $8x^2 + 3x + 1 < 0$ has no solution in terms of real $x$. The domain of $x$ satisfying this is an empty set.

Therefore, the solution is No solution.