Find x, if 9×3x=(27)2x−39 \times 3^x = (27)^{2x-3}9×3x=(27)2x−3
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9×3x=(27)2x−3⇒32×3x=(33)2x−3⇒32×3x=(3)3(2x−3)⇒32×3x=(3)6x−9⇒32=(3)6x−9÷3x⇒32=(3)6x−9−x [∵am÷an=am−n]⇒32=(3)5x−9⇒32=35x÷39⇒32×39=35x⇒32+9=35x [∵am×an=am+n]⇒311=35x⇒5x=11 [∵am=an⟹m=n]⇒x=115⇒x=2159 \times 3^x = (27)^{2x - 3}\\[1em] \Rightarrow 3^2 \times 3^x = (3^3)^{2x - 3}\\[1em] \Rightarrow 3^2 \times 3^x = (3)^{3(2x - 3)}\\[1em] \Rightarrow 3^2 \times 3^x = (3)^{6x - 9}\\[1em] \Rightarrow 3^2 = (3)^{6x - 9} ÷ 3^x\\[1em] \Rightarrow 3^2 = (3)^{6x - 9 - x} \space [∵a^m ÷ a^n = a^{m-n}] \\[1em] \Rightarrow 3^2 = (3)^{5x - 9}\\[1em] \Rightarrow 3^2 = 3^{5x} ÷ 3^9\\[1em] \Rightarrow 3^2 \times 3^9 = 3^{5x}\\[1em] \Rightarrow 3^{2 + 9} = 3^{5x} \space [∵a^m \times a^n = a^{m+n}] \\[1em] \Rightarrow 3^{11} = 3^{5x}\\[1em] \Rightarrow 5x = 11 \space [∵a^m = a^n ⟹m = n] \\[1em] \Rightarrow x = \dfrac{11}{5} \\[1em] \Rightarrow x = 2\dfrac{1}{5}9×3x=(27)2x−3⇒32×3x=(33)2x−3⇒32×3x=(3)3(2x−3)⇒32×3x=(3)6x−9⇒32=(3)6x−9÷3x⇒32=(3)6x−9−x [∵am÷an=am−n]⇒32=(3)5x−9⇒32=35x÷39⇒32×39=35x⇒32+9=35x [∵am×an=am+n]⇒311=35x⇒5x=11 [∵am=an⟹m=n]⇒x=511⇒x=251
If 9×3x=(27)2x−39 \times 3^x = (27)^{2x-3}9×3x=(27)2x−3, then x=215x = 2\dfrac{1}{5}x=251.
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