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प्रश्न
Find x, if
\[\left( \frac{- 1}{2} \right)^{- 19} \times \left( \frac{- 1}{2} \right)^8 = \left( \frac{- 1}{2} \right)^{- 2x + 1}\]
योग
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उत्तर
We have:
\[\left( \frac{- 1}{2} \right)^{- 19} \times \left( \frac{- 1}{2} \right)^8 = \left( \frac{- 1}{2} \right)^{- 2x + 1}\]
`((-1)/2)^(-11)=((-1)/2)^(-2x+1)`
\[( a^m \times a^n = a^{m + n} )\]
-11 = -2x + 1
-12 = -2x
6 = x
x = 6
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