The value of 25-28 - 2-7 is

The correct option is C 2^ 10Given, (2^5)/(2^8)× (2^ 7) =(2^5)× (2^ 7)/(2^8) ∵ (a^m) × (a^n) = (a^m+n) ⇒(2^5+( 7))/(2^8) = (2^5 7)/(2^8) = (2^ 2)/(2^8) ∵ (a^m ...

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

Byju's Answer

Standard VIII

Mathematics

Exponents with like Bases

The value of ...

Question

The value of $(\frac{2^{5}}{2^{8}}) \times 2^{- 7}$ is

2^{10}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2^{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2^{- 10}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2^{- 4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $2^{- 10}$
Given, $\frac{(2^{5})}{(2^{8})} \times (2^{- 7})$

= $\frac{(2^{5}) \times (2^{- 7})}{(2^{8})}$

$∵ (a^{m}) \times (a^{n}) = (a^{m + n})$

$\Rightarrow \frac{(2^{5 + (- 7)})}{(2^{8})}$ = $\frac{(2^{5 - 7})}{(2^{8})}$

= $\frac{(2^{- 2})}{(2^{8})}$
$∵ (a^{m}) \div (a^{n}) = (a^{m - n})$
= $2^{- 2 - 8}$
= $2^{- 10}$

Suggest Corrections

Similar questions

Q. The value of

\frac{(2^{5})}{(2^{8})} \times (2^{- 7})

Solve:-

$(\frac{2^{5}}{2^{8}}) \times 2^{- 7}$

Question 98(iv)

Simplify
$(2^{5} \div 2^{8}) \times 2^{- 7}$

Question 98(iv)

Simplify
$(2^{5} \div 2^{8}) \times 2^{- 7}$

Q. The value of

\frac{1}{2 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{10}} + . . . + \frac{1}{\sqrt{28} + \sqrt{31}}

is?

Join BYJU'S Learning Program

The value of (2528)×2−7 is

The correct option is C 2−10Given, (25)(28)× (2−7) =(25)× (2−7)(28) ∵(am)×(an) = (am+n) ⇒(25+(−7))(28) = (25−7)(28) = (2−2)(28) ∵(am)÷(an) = (am−n) = 2−2−8 = 2−10

The value of $(\frac{2^{5}}{2^{8}}) \times 2^{- 7}$ is