Question

Find the value of the following expression:

Answer 1

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

$\mathrm{Apply\:rule}\:a^0=1,\:a\ne \:0$

$6^0=1$

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

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

$\left(\frac{3^{-2}}{2^3}\right)^4=\frac{1^4}{\left(2^3\cdot \:3^2\right)^4}=\frac{1^4}{3^8\cdot \:2^{12}}$

$2^{28}\left(\frac{3^{-2}}{2^3}\right)^4\left(3^{-5}\cdot \:2^8\cdot \:1\right)^{-2}=\frac{243^2\cdot \:1\cdot \:2^{28}}{256^2\cdot \:3^8\cdot \:2^{12}}$

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

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

$=3^2$

$=9$