Question

State true or false.

${\left(\frac{16}{81}\right)}^{-\frac{3}{4}}\times {\left(\frac{49}{9}\right)}^{\frac{3}{2}}÷{\left(\frac{343}{216}\right)}^{\frac{2}{3}}$ is equal to $31.5$

• A. True
• B. False

Answer 1

The correct option is A True

As given ${\left(\frac{16}{81}\right)}^{-\frac{3}{4}}\times {\left(\frac{49}{9}\right)}^{\frac{3}{2}}÷{\left(\frac{343}{216}\right)}^{\frac{2}{3}}$

$\Rightarrow {\left(\frac{2^4}{3^4}\right)}^{-\frac{3}{4}}\times {\left(\frac{7^2}{3^2}\right)}^{\frac{3}{2}}÷{\left(\frac{7^3}{6^3}\right)}^{\frac{2}{3}}$

$\Rightarrow \left(\frac{2^{4\cdot -\frac{3}{4}}}{3^{4\cdot -\frac{3}{4}}}\right)\times \left(\frac{7^{2\cdot \frac{3}{2}}}{3^{2\cdot \frac{3}{2}}}\right)÷\left(\frac{7^{3\cdot \frac{2}{3}}}{6^{3\cdot \frac{2}{3}}}\right)$

$\Rightarrow \left(\frac{2^{-3}}{3^{-3}}\right)\times \left(\frac{7^3}{3^3}\right)÷\left(\frac{7^2}{6^2}\right)$

$\Rightarrow \left(\frac{2^{-3}}{3^{-3}}\right)\times \left(\frac{7^3}{3^3}\right)\times \left(\frac{6^2}{7^2}\right)$

$\Rightarrow \left(\frac{1\times 7^{3-2}\times 6^2}{2^3\times 3^{-3+3}}\right)$

$\Rightarrow \left(\frac{7\times 36}{8}\right)$

$\Rightarrow \frac{252}{8}$

$\Rightarrow 31.5$