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Serga [27]
3 years ago
9

(2^8 * 3^-5 * 6^0)^-2 * (3^-2/ 2^3) ^4 * 2^28

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
6 0

\left(2^8\cdot \:3^{-5}\cdot \:1\right)^{-2}=\left(1\cdot \frac{256}{243}\right)^{-2}=\left(\frac{256}{243}\right)^{-2}=\left(\frac{243}{256}\right)^2

\left(2^8\cdot \:3^{-5}\cdot \:6^0\right)^{-2}\left(\frac{3^{-2}}{2^3}\right)^4\cdot \:2^{28}

=2^{28}\left(\frac{3^{-2}}{2^3}\right)^4\left(3^{-5}\cdot \:2^8\cdot \:1\right)^{-2}

Consider

\left(2^8\cdot \:3^{-5}\cdot \:1\right)^{-2}=\left(1\cdot \frac{256}{243}\right)^{-2}=\left(\frac{243}{256}\right)^2=\frac{243^2}{256^2}

=2^{28}\left(\frac{3^{-2}}{2^3}\right)^4\frac{243^2}{256^2}

=2^{28}\cdot \frac{1}{2^{12}\cdot \:3^8}\cdot \frac{243^2}{256^2}

=\frac{243^2\cdot \:1\cdot \:2^{28}}{256^2\cdot \:3^8\cdot \:2^{12}}

=\frac{2^{16}\cdot \:243^2}{3^8\cdot \:256^2}

=\frac{2^{16}\cdot \:3^{10}}{2^{16}\cdot \:3^8}

=3^2

=9

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