Question

Find x, if
\[\left( \frac{- 1}{2} \right)^{- 19} \times \left( \frac{- 1}{2} \right)^8 = \left( \frac{- 1}{2} \right)^{- 2x + 1}\]

Answer 1

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