Find the value of x so that -23 - -2-6 - -22x-1

We have ( 2)^3 × ( 2)^ 6 = ( 2)^2x 1. We know that, a^m× a^n = a^m+n Now by using above law we can write given equation as, nbsp; ( 2)^3+( 6) = ( 2)^2x 1 ⇒ ...

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Standard VII

Mathematics

Zero Exponents

Find the valu...

Question

Find the value of x so that $(- 2)^{3} \times (- 2)^{- 6} = (- 2)^{2 x - 1}$

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Solution

We have $(- 2)^{3} \times (- 2)^{- 6} = (- 2)^{2 x - 1}$ .

We know that, $a^{m} \times a^{n} = a^{m + n}$

Now by using above law we can write given equation as,

$(- 2)^{3 + (- 6)} = (- 2)^{2 x - 1}$

$\Rightarrow (- 2)^{3 - 6} = (- 2)^{2 x - 1}$

$\Rightarrow (- 2)^{- 3} = (- 2)^{2 x - 1}$

Since, the base is same on both sides, then exponents must be the same.

$\Rightarrow - 3 = 2 x - 1 \Rightarrow 2 x = - 3 + 1$
$\Rightarrow 2 x = - 2$
$\Rightarrow x = - 1$
Hence $x = - 1$ will satisfy the given equation.

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Find the value of x so that (−2)3×(−2)−6=(−2)2x−1

Find the value of x so that $(- 2)^{3} \times (- 2)^{- 6} = (- 2)^{2 x - 1}$