Question

answer the both questions with explanation
wrongs and fun answers with be reported​

Answer 1

Answer :

Here, the concept of indices will be used. Before solving the question, let's have a look on the law of indices :

Law of indices :

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c}\\ \bf \: Law \: of \: Indices & \bf \: Examples \\ \\ \star \: \sf{1st \: Law \: (Product \: Law)} & \sf{{a}^{2} \times  {a}^{3} =  {a}^{2 + 3} = {a}^{5}  } \\ \\ \star \sf \: {2nd \: Law  \: (Quotient \: law) } & \sf{\dfrac{a^{2} }{a^{3}} = a^{2 - 3} = a^{ - 1}} \\ \\ \star \: \sf{3rd \: Law \: (Power \: law)} & \sf{(a^2)^3 = (a^{2 \times 3}) = (a^{6})} \end{array}}\end{gathered}\end{gathered}$

Question 1 :

Find the value of (70)⁻⁴ ÷ (70)⁻³

Solution :

Here, the quotient law will be used.

$\\ \longrightarrow \sf (70)^{-4} \div (70)^{-3}$

Here, the bases are equal, so we can subtract the powers. [2nd Law]

$\\ \longrightarrow \sf (70)^{-4 -(- 3)}$

$\\ \longrightarrow \sf (70)^{-4 + 3)}$

$\\ \longrightarrow \sf (70)^{-1}$

$\\ \longrightarrow \sf \dfrac{1}{70}$

$\boxed{\bf Answer ~ = ~ \dfrac{1}{70}}$

Question 2 :

Find the value of [5⁻¹ × 3⁻¹] ÷ 6⁻¹

Solution :

$\\ \longrightarrow \sf [5^{-1} \times 3^{-1}] \div 6^{-1}$

Here, the bases are not equal. So, we can't add the powers. So, firstly we will remove the negative power by reciprocating. And then we will do the required calculations.

$\\ \longrightarrow \sf \bigg[ \dfrac{1}{5} \times \dfrac{1}{3} \bigg] \div 6^{-1}$

$\\ \longrightarrow \sf \bigg[ \dfrac{1}{15} \bigg] \div 6^{-1}$

$\\ \longrightarrow \sf \bigg[ \dfrac{1}{15} \bigg] \div \dfrac{1}{6}$

$\\ \longrightarrow \sf \dfrac{1}{15} \times 6$

$\\ \longrightarrow \sf \dfrac{6}{15}$

$\\ \longrightarrow \sf \dfrac{ \not6}{ \not15}$

$\\ \longrightarrow \sf \dfrac{2}{5}$

$\boxed{\bf Answer ~ = ~ \dfrac{2}{5}}$

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Knowledge Bytes :

Signs are changed on the following bases :

• (-) × (-) = (+)
• (-) × (+) = (-)
• (+) × (+) = (+)
• (+) × (-) = (-)