Question

Question:- Solve this equation step by step

it's very urgent please solve now​

Answer 1

Answers :-

Question - 1 :- 74

Question - 2 :- 2¹³

Given to find the value of :-

$(2^2)^3+ \bigg(\dfrac{1}{3} \bigg)^0 +(3^2)^2\times\bigg\{\bigg(\dfrac{1}{3} \bigg)^{-1} \bigg\}^{-2}$

$\bigg\{\bigg(\dfrac{2}{3} \bigg)^{-2}+\bigg(\dfrac{4}{3} \bigg)^{-1} +(5)^3\bigg\}\div (2^3)^2$

Solution :-

Question - 1

$(2^2)^3+ \bigg(\dfrac{1}{3} \bigg)^0 +(3^2)^2\times\bigg\{\bigg(\dfrac{1}{3} \bigg)^{-1} \bigg\}^{-2}$

As we know some exponent laws that were ,

$(a^m)^n = a^{mn}$

$a^0 =1$

$\bigg(\dfrac{a}{b} \bigg)^{-n} = \bigg(\dfrac{b}{a} \bigg)^{n}$

$a^m\times a^n = a^{m+n}$

So, by using this We can solve this question

$2^{2\times3}+1 + 3^{2\times 2}\times (3^1)^{-2}$

$2^{6}+1 + 3^{4}\times (3)^{-2}$

$(64)+1 + 3^{4-2}$

$64+1+3^2$

$64+1+9$

$74$

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Question -2 :-

$\bigg\{\bigg(\dfrac{2}{3} \bigg)^{-2}+\bigg(\dfrac{4}{3} \bigg)^{-1} +(5)^3\bigg\}\div (2^3)^2$

Here also some exponent laws  were used that were ,

$\bigg(\dfrac{a}{b} \bigg)^{-n} = \bigg(\dfrac{b}{a} \bigg)^{n}$

$a^m \times a^n = a^{m+n}$

$(a^m)^n = a^{mn}$

$\bigg\{\bigg(\dfrac{3}{2} \bigg)^{2}+\bigg(\dfrac{3}{4} \bigg)^{1} +(5)^3\bigg\}\div (2^3)^2$

$\bigg\{\bigg(\dfrac{9}{4} \bigg)+\bigg(\dfrac{3}{4} \bigg) +125\bigg\}\div (2^6)$

$\bigg\{\dfrac{9}{4} +\dfrac{3}{4} +125\bigg\}\div (2^6)$

$\bigg\{\dfrac{9+3+4(125)}{4} \bigg\}\div2^6$

$\dfrac{12+500}{4}\times 2^6$

$\dfrac{512}{4} \times 2^6$

$\dfrac{2^9}{2^2} \times 2^6$

$\dfrac{2^9}{\not2^2} \times \not2^6 2^4$

$2^9 \times 2^4$ Since,

$a^m \times a^n = a^{m+n}$

$2^{9+4}$

$2^{13}$