Question

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

• A. \(log_2 \bigg(\frac{1}{3}\bigg)\)
• B. \(-log \bigg(\frac{1}{3}\bigg)\)
• C. \(log \bigg(\frac{1}{5}\bigg)\)
• D. \(-log \bigg(\frac{1}{5}\bigg)\)

Answer 1

The Correct Option is A

The correct option is (A): \(log_2 \bigg(\frac{1}{3}\bigg)\)

\(2^{y^2log_35} = log_23\)  

\((2log_35)^{y^2} = 5log_23\)  

\((5log_32)^{y^2} = 5log_23\)

\(5^{y^2log_32} = 5log_23\)

\(⇒ y^2log_32 = log_23\)

\(y^2 = (log_23)(log_23)\)

(is negative)

\(y = log_23^{-1}\) = \(log_2\frac{1}{3}\)

,

Answer 2

Given equation: \(2^{y^2log_3 5} = log_2 3\)
Using the property \(a^{bc} = (a^b)^c\): \((2log_3 5)^{y^2} = 5log_2 3\)
Given equation: \((5log_3 2)^{y^2} = 5log_2 3\)
Using the property \(a^{bc} = (a^c)^b\): \(5^{y^2log_3 2} = 5log_2 3\)
Equating the exponents: \(y^2log_3 2 = log_2 3\)
Simplifying the exponent: \(y^2 = (log_2 3)(log_2 3)\) (is negative)
Taking the inverse of \(log_2 3\) to get the final answer: \(y = log_2 3^{-1}\) = \(log_2 \frac{1}{3}\)
The correct option is (A): \(log_2 \left(\frac{1}{3}\right)\)