Question

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Answer 1

$\bf \begin{array}{clclll} 3^{1001}&\qquad &4^{1002}\\ \uparrow &&\uparrow \\ a&&b \end{array}\\\\ -----------------------------\\\\ (a+b)^2-(a-b)^2\implies (a^2+2ab+b^2)-(a^2-2ab+b^2) \\\\\\ a^2+2ab+b^2-a^2+2ab-b^2\implies 2ab+2ab\implies 4ab \\\\\\ now\qquad 4(3^{1001})(4^{1002})=k12^{1001}\qquad \begin{cases} 12^{1001}\\ (4\cdot 3)^{1001}\\ 4^{1001}\cdot 3^{1001} \end{cases}$

$\bf \\\\\\ 4(3^{1001})(4^{1002})=k(4^{1001}\cdot 3^{1001})\implies \cfrac{4(3^{1001})(4^{1002})}{4^{1001}\cdot 3^{1001}}=k \\\\\\ 4\cdot \cfrac{3^{1001}}{3^{1001}}\cdot \cfrac{4^{1002}}{4^{1001}}=k\implies 4\cdot 4^{1002}4^{-1001}=k\impliedby \begin{array}{llll} \textit{same base}\\ \textit{add the}\\ exponents \end{array} \\\\\\ 4\cdot 4^{1002-1001}=k\implies 4\cdot 4^1=k\implies 16=k$