Question

Verify given statements by solving both sides ----
1, ( - 7 )power32 ÷ ( - 7)power32 = 1​

Answer 1

Answer:

$(-7)^{32}\:\div(-7)^{32}\:=\:1$

Solving L.H.S :-

$(-7)^{32}\div(-7)^{32}$

= $(-7)^{32-32}$

= $(-7)^0$

= 1

Hence, L.H.S = R.H.S .

Answer 2

$\qquad \bigstar \ \large \overline{ \underline{ \frak{given}}}$

• (-7)³² ÷ (-7)³² = 1

$\qquad \bigstar \ \large \overline{ \underline{ \frak{to \: find}}}$

• Verify the given statement bu Solving both the side

$\qquad \bigstar \ \large \overline{ \underline{ \frak{solution}}}$

$\sf \ ∴\: \: \: L.H.S. = {( - 7)}^{32} \div {( - 7)}^{32} \\ \\ \qquad \sf \: = \frac{ {( - 7)}^{32} }{ {( - 7)}^{32} } \\ \\ \qquad \sf \: = \frac{ {( - 1)}^{32} {(7)}^{32} }{ {( - 1)}^{32} {(7)}^{32} }$

$\qquad \qquad\sf \: = \frac{1. {(7)}^{32} }{1. {(7)}^{32} }$

$\: \sf\qquad\qquad = \frac{ {(7)}^{32} }{ {(7)}^{32} } \: \: \: \: \: \: \: \: \: \bigg\{∵ \: \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n} \bigg\}$

$\sf \qquad \qquad = {(7)}^{32 - 32} \\ \\ \sf \qquad = {(7)}^{0} \\ \\\sf \qquad = 1$

$\sf \qquad \qquad \: = 1 = R. H. S. \\ \\ \qquad \sf \: So, L. H. S = R. H. S \qquad \qquad \qquad \qquad \frak{ \red{proved}}$