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$⇃ \underset{―}{}$

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((^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)

$\frac{^{3} \sqrt{4^{5 - x}}}{\int_{4}^{6} (x - 1) d x} = \frac{1}{2^{2 x - 1}}, find the value of x .$ $\underset{―}{Solution}$ $\frac{4^{\frac{5 - x}{3}}}{\int_{4}^{6} (\frac{x^{2}}{2} - x + k)} = \frac{1}{2^{2 x - 1}}$ $\frac{2^{2 ∙ \frac{5 - x}{3}}}{(\frac{6^{2}}{2} - 6 + k) - (\frac{4^{2}}{2} - 4 + k)} = \frac{1}{2^{2 x - 1}}$ $\frac{2^{\frac{10 - 2 x}{3}}}{\frac{36}{2} - 6 + k - \frac{16}{2} + 4 - k} = \frac{1}{2^{2 x - 1}}$ $\frac{2^{\frac{10 - 2 x}{3}}}{18 - 6 - 8 + 4} = \frac{1}{2^{2 x - 1}}$ $\frac{2^{\frac{10 - 2 x}{3}}}{8} = \frac{1}{2^{2 x - 1}} (Cross Multiply)$ $2^{2 x - 1} \times 2^{\frac{10 - 2 x}{3}} = 8 \times 1$ $2^{2 x - 1} \times 2^{\frac{10 - 2 x}{3}} = 2^{3}$ $2^{2 x - 1 + \frac{10 - 2 x}{3}} = 2^{3} (Since, the bases are equal . Then, we can equate the exponents)$ $2 x - 1 + \frac{10 - 2 x}{3} = 3 (Multiply each term by 3)$ $3 (2 x) - 3 (1) + 3 (\frac{10 - 2 x}{3}) = 3 (3)$ $6 x - 3 + 10 - 2 x = 9 (Collect Like Terms)$ $4 x + 7 = 9$ $4 x = 9 - 7$ $4 x = 2 (Divide Both Sides by 4)$ $\frac{4 x}{4} = \frac{2}{4}$ $∴ x = \frac{1}{2}$