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Question Number 202636 by ibroclex_adex last updated on 30/Dec/23

((^3 (√4^(5−x) ))/(∫_4 ^6 (x−1)dx)) = (1/2^(2x−1) ) , find the value of x. Solution (4^((5−x)/3) /(∫_4 ^6 ((x^2 /2)−x+k))) = (1/2^(2x−1) ) (2^(2•((5−x)/3)) /(((6^2 /2)−6+k)−((4^2 /2)−4+k))) = (1/2^(2x−1) ) (2^((10−2x)/3) /(((36)/2)−6+k−((16)/2)+4−k)) = (1/2^(2x−1) ) (2^((10−2x)/3) /(18−6−8+4)) = (1/2^(2x−1) ) (2^((10−2x)/3) /8) = (1/2^(2x−1) ) (Cross Multiply) 2^(2x−1) ×2^((10−2x)/3) = 8×1 2^(2x−1) ×2^((10−2x)/3) = 2^3 2^(2x−1+((10−2x)/3)) = 2^3 (Since, the bases are equal. Then, we can equate the exponents) 2x−1+((10−2x)/3) = 3 (Multiply each term by 3) 3(2x)−3(1)+3(((10−2x)/3)) = 3(3) 6x−3+10−2x = 9 (Collect Like Terms) 4x+7 = 9 4x = 9−7 4x = 2 (Divide Both Sides by 4) ((4x)/4) = (2/4) ∴ x = (1/2)

345−x∫46(x−1)dx=122x−1,findthevalueofx.Solution―45−x3∫46(x22−x+k)=122x−122∙5−x3(622−6+k)−(422−4+k)=122x−1210−2x3362−6+k−162+4−k=122x−1210−2x318−6−8+4=122x−1210−2x38=122x−1(CrossMultiply)22x−1×210−2x3=8×122x−1×210−2x3=2322x−1+10−2x3=23(Since,thebasesareequal.Then,wecanequatetheexponents)2x−1+10−2x3=3(Multiplyeachtermby3)3(2x)−3(1)+3(10−2x3)=3(3)6x−3+10−2x=9(CollectLikeTerms)4x+7=94x=9−74x=2(DivideBothSidesby4)4x4=24∴x=12

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