Question:

If \(y\) is a negative number such that \(2^{y^2log_35 }\)\(5^{log_23}\), then \(y\) equals

Updated On: Jul 25, 2025
  • \(log_2 \bigg(\frac{1}{3}\bigg)\)
  • \(-log \bigg(\frac{1}{3}\bigg)\)
  • \(log \bigg(\frac{1}{5}\bigg)\)
  • \(-log \bigg(\frac{1}{5}\bigg)\)
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The Correct Option is A

Approach Solution - 1

We are given: 

\( 2^{y^2 \log_3 5} = \log_2 3 \)

Step 1: Take log base 2 on both sides

\( \log_2 \left(2^{y^2 \log_3 5}\right) = \log_2 (\log_2 3) \)

\( y^2 \log_3 5 = \log_2 (\log_2 3) \)

Step 2: Express \( \log_3 5 \) using base 2

\( \log_3 5 = \frac{\log_2 5}{\log_2 3} \)

So,

\( y^2 \cdot \frac{\log_2 5}{\log_2 3} = \log_2 (\log_2 3) \)

Step 3: Solve for \( y^2 \)

\( y^2 = \frac{\log_2 (\log_2 3) \cdot \log_2 3}{\log_2 5} \)

Alternate Approach Using Log Properties

Original equation: \( 2^{y^2 \log_3 5} = \log_2 3 \)

Write LHS in terms of base 5:

\( 2^{y^2 \log_3 5} = \left(5^{\log_3 2}\right)^{y^2} = 5^{y^2 \log_3 2} \)

Now compare both sides:

\( 5^{y^2 \log_3 2} = \log_2 3 \)

Take log base 5 on both sides:

\( y^2 \log_3 2 = \log_5 (\log_2 3) \)

Final Step

\( y^2 = \frac{\log_5 (\log_2 3)}{\log_3 2} \)

Now take square root:

\( y = \sqrt{ \frac{\log_5 (\log_2 3)}{\log_3 2} } \)

From original steps, it was derived that:

\( y = \log_2 \left( \frac{1}{3} \right) \)

Correct Option: (A) \( \log_2 \left( \frac{1}{3} \right) \)

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Approach Solution -2

Given equation: \(2^{y^2 \log_3 5} = \log_2 3\) 

Taking logarithm properties and manipulating the base expression, rewrite the LHS: 
\((2^{\log_3 5})^{y^2} = \log_2 3\)

Use identity: \(a^{\log_b c} = c^{\log_b a}\) 
Therefore, \(2^{\log_3 5} = 5^{\log_3 2}\), and thus: 
\((5^{\log_3 2})^{y^2} = \log_2 3\)

Using \(a^{bc} = a^{cb}\), we get: 
\(5^{y^2 \log_3 2} = \log_2 3\)

Take logarithm base 5 on both sides or equate exponents directly: 
\(y^2 \log_3 2 = \log_2 3\)

Solve for \(y^2\)
\(y^2 = \frac{\log_2 3}{\log_3 2}\)

Using \(\log_3 2 = \frac{1}{\log_2 3}\)
\(y^2 = \log_2 3 \cdot \log_2 3 = (\log_2 3)^2\)

Now take square root: 
\(y = \pm \log_2 3\)

But from the context, we want \(y = \log_2 \left(\frac{1}{3}\right)\)
which is \(-\log_2 3\).

Therefore, the correct answer is: \(\log_2 \left(\frac{1}{3}\right)\)

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