Choose the statements that are true.

2^{2} \times 2^{3} = 2^{5}

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

\frac{3^{6}}{3^{2}} = 3^{4}

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

3^{3} \times 3^{3} = 3^{9}

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

\frac{2^{6}}{2^{2}} = 2^{3}

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

Open in App

Solution

The correct options are
A $2^{2} \times 2^{3} = 2^{5}$
B $\frac{3^{6}}{3^{2}} = 3^{4}$
Using laws of exponents, $a^{m} \times a^{n} = a^{m + n}$ and $\frac{a^{m}}{a^{n}} = a^{m - n} .$

$∴ 2^{2} \times 2^{3} = 2^{2 + 3} = 2^{5}$
$\frac{3^{6}}{3^{2}} = 3^{6 - 2} = 3^{4}$
$3^{3} \times 3^{3} = 3^{3 + 3} = 3^{6}$
$\frac{2^{6}}{2^{2}} = 2^{6 - 2} = 2^{4}$

Suggest Corrections

Similar questions

2^{3} \times 3^{3} \times 7^{2}

b = 2^{2} \times 3^{2} \times 5^{3},

lim n \to \infty \frac{n (1^{2} + 2^{2} + 3^{2} + . . \cdot . \cdot . \cdot \cdot + n^{2})^{5}}{(1^{3} + 2^{3} + 3^{3} + + n^{3})^{4}} =

\frac{3 \times 1^{3}}{1^{2}} + \frac{5 \times (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 \times (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + \dots

10^{th}

\frac{1^{3} + 2^{3} + 3^{3} + 4^{3} + . . . . . . . {.12}^{3}}{1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . . {.12}^{2}} =

\frac{3 \times 1^{3}}{1^{2}} + \frac{5 \times (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 \times (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + \dots upto 10 terms

Join BYJU'S Learning Program