Answered

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How to solve :

〖1/(3^(x-2).9^(3x) )=1/(〖√(〖27〗^(2x+3) ))〗^(-1) )〗^

Sagot :

Konrad509idn
[tex]\dfrac{1}{3^{x-2}\cdot9^{3x}}=\dfrac{1}{(\sqrt{27^{2x+3}})^{-1}}\\ \dfrac{1}{3^{x-2}\cdot(3^2)^{3x}}=\dfrac{1}{(((3^3)^{2x+3})^{\frac{1}{2}})^{-1}}\\ \dfrac{1}{3^{x-2}\cdot3^{6x}}=\dfrac{1}{((3^{6x+9})^{\frac{1}{2}})^{-1}}\\ \dfrac{1}{3^{x-2+6x}}=\dfrac{1}{(3^{3x+\frac{9}{2}})^{-1}}\\ \dfrac{1}{3^{7x-2}}=\dfrac{1}{3^{-3x-\frac{9}{2}}}\\ 3^{7x-2}=3^{-3x-\frac{9}{2}}\\ 7x-2=-3x-\dfrac{9}{2}\\ 14x-4=-6x-9\\ 20x=-5\\ x=-4 [/tex]
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