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): $log_2 \left(\frac{1}{3}\right)$