Question 99ii-23 - -26 - -22x-1

We have ( 2)^3 × ( 2)^6 = ( 2)^2x 1 Using law of exponents, a^ m = 1/a^m Then,( 2)^3( 2)^ 6 = ( 2)^2x 1 ⇒ nbsp;( 2)^3+( 6) = ( 2)^2x 1 ⇒ nbsp;( 2)^3 6 = ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VII

Mathematics

Exponent of a Product

Question 99ii...

Question

Question 99(ii)

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

Open in App

Solution

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

Using law of exponents,
$a^{- m} = \frac{1}{a^{m}}$

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

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

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

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

Since, the base is same on both sides, we can equate the exponents.

$- 3 = 2 x - 1 \Rightarrow 2 x = - 3 + 1$
$\Rightarrow 2 x = - 2$
$\Rightarrow x = - 1$

Suggest Corrections

Similar questions

Q. Solve for

x

2^{2 x + 3} + 2^{5} = 2^{6}

\sqrt{3} x^{2} + 2 \sqrt{2} x - 2 \sqrt{3} = 0

Q. Find the value of x so that

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

Q. Find the value of x if

2^{6} \times 2^{2 x} = 2^{10}

Q. evaluate

\frac{{(\frac{1}{2})}^{5} - {(\frac{1}{2})}^{6}}{{(\frac{1}{2})}^{3} - {(\frac{1}{2})}^{5}}

Join BYJU'S Learning Program

Explore more

Exponent of a Product

Standard VII Mathematics

Join BYJU'S Learning Program

Question 99(ii) (−2)3×(−2)6=(−2)2x−1

Question 99(ii)

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