Question

Find the value of x, if 3 + 2x = 64 ^1/2 + 27^1/3
Please answer fast
I will mark the best answer as brainliest

Answer 1

$\bf Given : \rm 3+2x=(64)^{\frac{1}{2}}+(27)^{\frac{1}{3}}$

$\bf To~ find : \rm The ~value~ of ~x.$

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$\rm We~ can~ write~ the ~ numbers~ (64)^{\frac{1}{2}}~and~ (27)^{\frac{1}{3}}~ in~ more~ simplified~ form:$

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$\rm \qquad\dashrightarrow~ (64)^{\frac{1}{2}}=(8^2)^{\frac{1}{2}}$

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$\rm Using~ the~ law:(a^m)^n=(a)^{mn}$

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$\qquad\rm\dashrightarrow~(8^{\cancel2})^{\frac{1}{\cancel2}}= \bf8$

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$\qquad\rm\dashrightarrow~ (27)^{\frac{1}{3}}=(3^3)^{\frac{1}{3}}$

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$\rm Again~ using~ the~ law:(a^m)^n=(a)^{mn}$

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$\rm\qquad\dashrightarrow~ (3^{\cancel3})^{\frac{1}{\cancel3}}=\bf 3$

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$\rm Now~substituting~ and~ solving:$

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$\rm\qquad\dashrightarrow~ 3+2x=8+3$

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$\qquad\rm\dashrightarrow~ 3+2x=11$

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$\qquad\rm\dashrightarrow~ 2x=11-3$

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$\rm\qquad\dashrightarrow~ 2x=8$

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$\rm\qquad\dashrightarrow~ x=\cancel{\dfrac{8}{2} }$

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$\bf\qquad\dashrightarrow~ x=4$

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$\underline{\rm Hence, ~the~value~ of~ x~in~3+2x=(64)^{\frac{1}{3}}+(27)^{\frac{1}{3}}~is~\bf4.}$

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