Question

find the value if 81 upon 16 raise to power minus 3 upon 4 into 25 upon 9 whole raise to power minus 3 upon 2

Answer 1

This is your answer
May this will help you

Answer 2

Answer:

$\text{The value of }(\frac{81}{16})^{\frac{-3}{4}}\times (\frac{25}{9})^{\frac{-3}{2}}\text{ is }\frac{8}{125}$  

Step-by-step explanation:

$\text{we have to find the value of }(\frac{81}{16})^{\frac{-3}{4}}\times (\frac{25}{9})^{\frac{-3}{2}}$

$(\frac{81}{16})^{\frac{-3}{4}\times (\frac{25}{9})^{\frac{-3}{2}}$

The above expression can be written as

$(\frac{3^4}{2^4})^{\frac{-3}{4}}\times (\frac{5^2}{3^2})^{\frac{-3}{2}}$

$(\frac{3}{2})^{4\times {\frac{-3}{4}}}\times (\frac{5}{3})^{2\times {\frac{-3}{2}}$

$(\frac{3}{2})^{-3} \times (\frac{5}{3})^{-3}$

$\text{As }x^{-a}=\frac{1}{x^a}$

$(\frac{2}{3})^{3} \times (\frac{3}{5})^{3}$

$\text{As }x^{a}y^a=(xy)^a$

$(\frac{2}{3}\times \frac{3}{5})^3$

$(\frac{2}{5})^3=\frac{8}{125}$

$\text{Hence, the value of }(\frac{81}{16})^{\frac{-3}{4}}\times (\frac{25}{9})^{\frac{-3}{2}}\text{ is }\frac{8}{125}$