Solve :
{(625−12)−14}2=(0.2)4−3x\Big{\Big(625^{-\dfrac{1}{2}}\Big)^{-\dfrac{1}{4}}\Big}^2 = (0.2)^{4-3x}{(625−21)−41}2=(0.2)4−3x
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{(625−12)−14}2=(0.2)4−3x⇒{(625−12)−24}=(0.2)4−3x⇒{(625−12)−12}=(0.2)4−3x⇒{(625−12×−12)}=(0.2)4−3x⇒62514=(0.2)4−3x⇒(54)14=(0.2)4−3x⇒5=(0.2)4−3x⇒5=2104−3x⇒5=154−3x⇒5=53x−4⇒3x−4=1⇒3x=1+4⇒3x=5⇒x=53=123\Big{\Big(625^{-\dfrac{1}{2}}\Big)^{-\dfrac{1}{4}}\Big}^2 = (0.2)^{4-3x}\\[1em] \Rightarrow \Big{\Big(625^{-\dfrac{1}{2}}\Big)^{-\dfrac{2}{4}}\Big} = (0.2)^{4-3x}\\[1em] \Rightarrow \Big{\Big(625^{-\dfrac{1}{2}}\Big)^{-\dfrac{1}{2}}\Big} = (0.2)^{4-3x}\\[1em] \Rightarrow \Big{\Big(625^{-\dfrac{1}{2} \times -\dfrac{1}{2}}\Big)\Big} = (0.2)^{4-3x}\\[1em] \Rightarrow 625^{\dfrac{1}{4}} = (0.2)^{4-3x}\\[1em] \Rightarrow {(5^4)}^{\dfrac{1}{4}} = (0.2)^{4-3x}\\[1em] \Rightarrow 5 = (0.2)^{4-3x}\\[1em] \Rightarrow 5 = \dfrac{2}{10}^{4-3x}\\[1em] \Rightarrow 5 = \dfrac{1}{5}^{4-3x}\\[1em] \Rightarrow 5 = 5^{3x-4}\\[1em] \Rightarrow 3x - 4 = 1\\[1em] \Rightarrow 3x = 1 + 4\\[1em] \Rightarrow 3x = 5\\[1em] \Rightarrow x = \dfrac{5}{3} = 1\dfrac{2}{3}{(625−21)−41}2=(0.2)4−3x⇒{(625−21)−42}=(0.2)4−3x⇒{(625−21)−21}=(0.2)4−3x⇒{(625−21×−21)}=(0.2)4−3x⇒62541=(0.2)4−3x⇒(54)41=(0.2)4−3x⇒5=(0.2)4−3x⇒5=1024−3x⇒5=514−3x⇒5=53x−4⇒3x−4=1⇒3x=1+4⇒3x=5⇒x=35=132
Hence, the value of x = 1231\dfrac{2}{3}132.
Answered By
Beaker A contains sugar solution with 18 percent sugar.beaker B contains sugar solution with 12 percent sugar.How much of each must he mixed together to get solution of 16 percent sugar weighing 240 gm of it ?
Solve for x :
320+23=(0.6)2−3x\sqrt{32^0+\dfrac{2}{3}} = (0.6)^{2-3x}320+32=(0.6)2−3x.
Evaluate :
xa−b×xb−c×xc−a\sqrt{x^{a-b}} \times \sqrt{x^{b-c}} \times \sqrt{x^{c-a}}xa−b×xb−c×xc−a
Simplify:
(x+1y)a(x−1y)b(y+1x)a(y−1x)b\dfrac{\Big(x+\dfrac{1}{y}\Big)^a\Big(x-\dfrac{1}{y}\Big)^b}{\Big(y+\dfrac{1}{x}\Big)^a\Big(y-\dfrac{1}{x}\Big)^b}(y+x1)a(y−x1)b(x+y1)a(x−y1)b