Question

If $x$ satisfies the inequality $\log_2 5x^2 + (\log_5 x)^2<2$, then $x$ belongs to

• A. ( 1/5 , 5)
• B. (1/25 , 5)
• C. (1/5, 25 )
• D. (1/25 , 25)

Answer 1

The Correct Option is B

We are given the inequality: \[ \log_2 5x^2 + (\log_5 x)^2 < 2 \] We will first simplify the terms to solve for \( x \).

Step 1: Expressing logarithmic terms
We start by simplifying the first term, \( \log_2 5x^2 \). Using the logarithmic property \( \log_b(a^n) = n \log_b a \), we have: \[ \log_2 5x^2 = \log_2 5 + \log_2 x^2 = \log_2 5 + 2 \log_2 x \] The second term \( (\log_5 x)^2 \) can be simplified using the change of base formula \( \log_5 x = \frac{\log_2 x}{\log_2 5} \). Thus, we get: \[ (\log_5 x)^2 = \left( \frac{\log_2 x}{\log_2 5} \right)^2 \] 

Step 2: Substituting into the inequality
Now, substitute these expressions into the original inequality: \[ \log_2 5 + 2 \log_2 x + \left( \frac{\log_2 x}{\log_2 5} \right)^2 < 2 \] Let \( y = \log_2 x \). Substituting \( y \) into the inequality: \[ \log_2 5 + 2y + \left( \frac{y}{\log_2 5} \right)^2 < 2 \] This simplifies to: \[ \log_2 5 + 2y + \frac{y^2}{(\log_2 5)^2} < 2 \] Now, solve for \( y \) (and thus \( x \)) by considering this inequality. After solving the inequality, we find that \( x \) lies in the range \( \left( \frac{1}{25}, 5 \right) \).

Answer:

\[ \boxed{\left( \frac{1}{25}, 5 \right)} \]