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

2 thoughts on “Verify given statements by solving both sides —- <br />1, ( – 7 )power32 ÷ ( – 7)power32 = 1​”

1. [tex] \qquad \bigstar \ \large \overline{ \underline{ \frak{given}}}[/tex]

   • (-7)³² ÷ (-7)³² = 1

   [tex] \qquad \bigstar \ \large \overline{ \underline{ \frak{to \: find}}}[/tex]

   • Verify the given statement bu Solving both the side

   [tex] \qquad \bigstar \ \large \overline{ \underline{ \frak{solution}}}[/tex]

   [tex] \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} } \\ \\ [/tex]

   [tex] \qquad \qquad\sf \: = \frac{1. {(7)}^{32} }{1. {(7)}^{32} } \\ \\ [/tex]

   [tex] \: \sf\qquad\qquad = \frac{ {(7)}^{32} }{ {(7)}^{32} } \: \: \: \: \: \: \: \: \: \bigg\{∵ \: \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m – n} \bigg\} \\ \\ [/tex]

   [tex] \sf \qquad \qquad = {(7)}^{32 – 32} \\ \\ \sf \qquad = {(7)}^{0} \\ \\\sf \qquad = 1 \\\\ [/tex]

   [tex] \sf \qquad \qquad \: = 1 = R. H. S. \\ \\ \qquad \sf \: So, L. H. S = R. H. S \qquad \qquad \qquad \qquad \frak{ \red{proved}}[/tex]

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2. Answer:

   [tex](-7)^{32}\:\div(-7)^{32}\:=\:1[/tex]

   Solving L.H.S :–

   [tex](-7)^{32}\div(-7)^{32}[/tex]

   = [tex](-7)^{32-32}[/tex]

   = [tex](-7)^0[/tex]

   = 1

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

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