Find the value of $ K $ in $ \left(\frac{3}{5}\right)^{3} \times\left(\frac{3}{5}\right)^{-6}=\left(\frac{5}{3}\right)^{1-2 k} $

Given: $\left(\frac{3}{5}\right)^{3} \ \times \ \left(\frac{3}{5}\right)^{-6} \ =\ \left(\frac{5}{3}\right)^{1-2k}$

To find: We have to find the value of k.

Solution: 

$\left(\frac{3}{5}\right)^{3} \ \times \ \left(\frac{3}{5}\right)^{-6} \ =\ \left(\frac{5}{3}\right)^{1-2k}$

$\left(\frac{3}{5}\right)^{3-6} \ =\ \left(\frac{5}{3}\right)^{1-2k}$

$\left(\frac{3}{5}\right)^{-3} \ =\ \left(\frac{5}{3}\right)^{1-2k}$

$\left(\frac{5}{3}\right)^{3} \ =\ \left(\frac{5}{3}\right)^{1-2k}$

Since, bases are equal the exponents must be equal.

Therefore, $1-2k = 3$

=> $-2 = 2k$

=> $k = -1$.

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Updated on: 10-Oct-2022

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