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Natali [406]
3 years ago
15

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

Mathematics
1 answer:
grin007 [14]3 years ago
6 0
\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

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