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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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