Find $ y $, if
$ \left(-\frac{1}{2}\right)^{-20} \p\left(-\frac{1}{2}\right)^{9}=\left(-\frac{1}{2}\right)^{-2 y+1} $


Given:

\( \left(-\frac{1}{2}\right)^{-20} \div\left(-\frac{1}{2}\right)^{9}=\left(-\frac{1}{2}\right)^{-2 y+1} \)

To do:

We have to find the value of $y$.

Solution:

We know that,

$a^m \div a^n=a^{(m-n)}$

Therefore,

$(-\frac{1}{2})^{-20} \div(-\frac{1}{2})^{9}=(-\frac{1}{2})^{-2 y+1}$

$(-\frac{1}{2})^{(-20-9)}=(-\frac{1}{2})^{-2 y+1}$

$(-\frac{1}{2})^{-29}=(-\frac{1}{2})^{-2 y+1}$

Comparing the powers on both sides, we get,

$-29=-2y+1$

$2y=1+29$

$2y=30$

$y=\frac{30}{2}$

$y=15$

The value of $y$ is $15$.

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Updated on: 10-Oct-2022

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