If left(frac{x}{y}right)^{5a-3} = left(frac{y | vedclass

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Question

(A) Given the equation: $\left(\frac{x}{y}\right)^{5a-3} = \left(\frac{y}{x}\right)^{17-3a}$.We know that $\frac{y}{x} = \left(\frac{x}{y}\right)^{-1}

Vedclass · Accepted Answer

$-7$

Vedclass · Answer

(A) Given the equation: $\left(\frac{x}{y}\right)^{5a-3} = \left(\frac{y}{x}\right)^{17-3a}$.We know that $\frac{y}{x} = \left(\frac{x}{y}\right)^{-1}$.Substituting this into the right side of the equation:$\left(\frac{x}{y}\right)^{5a-3} = \left(\left(\frac{x}{y}\right)^{-1}\right)^{17-3a}$.Using the power of a power rule $(x^m)^n = x^{mn}$:$\left(\frac{x}{y}\right)^{5a-3} = \left(\frac{x}{y}\right)^{-(17-3a)}$.$\left(\frac{x}{y}\right)^{5a-3} = \left(\frac{x}{y}\right)^{3a-17}$.Since the bases are equal,we equate the exponents:$5a - 3 = 3a - 17$.Subtract $3a$ from both sides:$2a - 3 = -17$.Add $3$ to both sides:$2a = -14$.Divide by $2$:$a = -7$.

If $\left(\frac{x}{y}\right)^{5a-3} = \left(\frac{y}{x}\right)^{17-3a}$,what is the value of $a$?

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