Change-of-Base Formula for Logarithms: Inequality

To solve this problem, we will rewrite both sides of the inequality with the change of base formula and evaluate them:

• Step 1: Rewrite both logarithms using common base.
  Using the change of base formula, $\log_{\frac{1}{4}}9 = \frac{\log_29}{\log_2\frac{1}{4}}$ and $\frac{\log_57}{\log_5\frac{1}{4}}$ can also be rewritten similarly.
• Step 2: Recognize that changing everything to base 2 will be beneficial.
  $\log_2 (1/4) = \log_2 4^{-1} = -2$.
• Step 3: Evaluate left side.
  $\log_{\frac{1}{4}}9 = \frac{\log_2 9}{-2} = -\frac{\log_2 9}{2}$.
• Step 4: Evaluate right side using the same base.
  $\frac{\log_57}{\log_5\frac{1}{4}} = -\frac{\log_2 7}{2}$, where $-\log_2 4 = 2$.
• Step 5: Compare both expressions
  $-\frac{\log_2 9}{2} < -\frac{\log_2 7}{2} => \log_2 9 < \log_2 7 => 9 < 7$.
• Step 6: Since $9 > 7$, convert the logarithmic expressions back into $\log_{\frac{1}{4}}$.
  $\log_{\frac{1}{4}}7$ is smaller than $\log_{\frac{1}{4}}9$, so inequation holds.

After comparing these expressions, we see that $\log_{\frac{1}{4}}9 < \log_{\frac{1}{4}}7$ indeed holds true.

Therefore, the solution is: Yes, since: $\log_{\frac{1}{4}}9 < \log_{\frac{1}{4}}7$.

To solve the inequality $\frac{\log_{\frac{1}{8}}(2x)}{\log_{\frac{1}{8}}(4)} < \log_4(5x - 2)$, we proceed as follows:

• Step 1: Convert all logarithms to a common base using the change of base formula:

  $\log_{\frac{1}{8}}(a) = \frac{\log(a)}{\log(\frac{1}{8})}$ and $\log_4(b) = \frac{\log(b)}{\log(4)}$.

• Step 2: Simplify the inequality using these conversions.

  The left expression becomes $\frac{\log(2x)}{\log(\frac{1}{8})} \div \frac{\log(4)}{\log(\frac{1}{8})} = \frac{\log(2x)}{\log(4)}$.

• Step 3: The inequality simplifies to $\frac{\log(2x)}{\log(4)} < \frac{\log(5x - 2)}{\log(4)}$.
• Step 4: Since both sides are divided by the positive $\log(4)$, the inequality remains:

  $\log(2x) < \log(5x - 2)$.

• Step 5: Remove logs since the logarithms are to the same base, leading to $2x < 5x - 2$.
• Step 6: Solve the inequality $2x < 5x - 2$. Rearrange terms: $2 < 3x$.
• Step 7: Divide both sides by 3 to solve for $x$: $\frac{2}{3} < x$.
• Step 8: Validate $(5x - 2) > 0$ implies $x > \frac{2}{5}$, which is consistent with our solution.

Therefore, the solution to the problem is $\frac{2}{3} < x$, which is choice 1.

To solve the problem:

• Given the inequality $\frac{\log_3(x^2+5x+4)}{\log_3x}<\log_x12$, apply the change-of-base formula:
• Rewrite $\log_x 12 = \frac{\log_3 12}{\log_3 x}$.
• Substitute into the inequality:
• $\frac{\log_3(x^2 + 5x + 4)}{\log_3 x} < \frac{\log_3 12}{\log_3 x}$.
• Cross multiply assuming $\log_3 x > 0$ (because $x > 1$): $\log_3(x^2 + 5x + 4) < \log_3 12$.
• The inequality $\log_3(x^2 + 5x + 4) < \log_3 12$ implies:
• $x^2 + 5x + 4 < 12$.
• Simplify: $x^2 + 5x + 4 - 12 < 0$, which gives $x^2 + 5x - 8 < 0$.
• Find roots for $x^2 + 5x - 8 = 0$ using the quadratic formula:
• $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1, b = 5, c = -8$.
• $x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-8)}}{2}$.
• $x = \frac{-5 \pm \sqrt{57}}{2}$.
• We find the intervals projecting on the inequality sign: $x^2 + 5x - 8 < 0$.
• Analyzing the sign change for $1 < x < \frac{-5 + \sqrt{57}}{2}$.
• Additionally, confirm $x^2 + 5x + 4 > 0$ for valid logarithm argument, which is naturally satisfied in previous constraints.

Therefore, the solution is: $\mathbf{1 < x < -2.5+\frac{\sqrt{57}}{2}}$.