Solve Logarithmic Inequality: log₁/₄(9) vs log₅(7)/log₅(1/4)

Is inequality true?

$\log_{\frac{1}{4}}9<\frac{\log_57}{\log_5\frac{1}{4}}$

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: $\log_{\frac{1}{4}}9 < \log_{\frac{1}{4}}7$.