Given (827)x−1\Big(\dfrac{8}{27}\Big)^{x-1}(278)x−1 = (94)2x+1\Big(\dfrac{9}{4}\Big)^{2x+1}(49)2x+1; find the value of x .
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(827)x−1=(94)2x+1⇒(2333)x−1=(3222)2x+1⇒(23)3(x−1)=(32)2(2x+1)⇒(23)3(x−1)=(23)−2(2x+1)⇒3(x−1)=−2(2x+1)⇒3x−3=−4x−2⇒3x+4x=3−2⇒7x=1⇒x=17\Big(\dfrac{8}{27}\Big)^{x-1} = \Big(\dfrac{9}{4}\Big)^{2x+1}\\[1em] ⇒ \Big(\dfrac{2^3}{3^3}\Big)^{x-1} = \Big(\dfrac{3^2}{2^2}\Big)^{2x+1}\\[1em] ⇒ \Big(\dfrac{2}{3}\Big)^{3(x-1)} = \Big(\dfrac{3}{2}\Big)^{2(2x+1)}\\[1em] ⇒ \Big(\dfrac{2}{3}\Big)^{3(x-1)} = \Big(\dfrac{2}{3}\Big)^{-2(2x+1)}\\[1em] ⇒ 3(x - 1) = -2(2x + 1)\\[1em] ⇒ 3x - 3 = -4x - 2\\[1em] ⇒ 3x + 4x = 3 - 2\\[1em] ⇒ 7x = 1\\[1em] ⇒ x = \dfrac{1}{7}\\[1em](278)x−1=(49)2x+1⇒(3323)x−1=(2232)2x+1⇒(32)3(x−1)=(23)2(2x+1)⇒(32)3(x−1)=(32)−2(2x+1)⇒3(x−1)=−2(2x+1)⇒3x−3=−4x−2⇒3x+4x=3−2⇒7x=1⇒x=71
Hence, the value of x = 17\dfrac{1}{7}71.
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If 25x+1=1255x25^{x+1} = \dfrac{125}{5^x}25x+1=5x125 ; find the value of x .
If 8x×4y=328^x \times 4^y = 328x×4y=32 and 81x÷27y=381^x \div 27^y = 381x÷27y=3; find the values of x and y .
Evaluate:
14+(0.01)−12×(5)−(27)23\sqrt\dfrac{1}{4} + (0.01)^{-\dfrac{1}{2}} \times (5) - (27)^\dfrac{2}{3}41+(0.01)−21×(5)−(27)32
(14)−2−3(32)25×(7)0+(916)−12\Big(\dfrac{1}{4}\Big)^{-2} - 3(32)^\dfrac{2}{5}\times (7)^0 + \Big(\dfrac{9}{16}\Big)^{-\dfrac{1}{2}}(41)−2−3(32)52×(7)0+(169)−21
