Question

Answer This question

Answer 1

Answer:

1)

a=3

2)

$\dfrac{1}{27}$

Step-by-step explanation:

Use the property of exponent, $a^m\cdot a^n=a^{m+n}$

$\left(\dfrac{2}{3}\right)^{-3+8}=\left(\dfrac{2}{3}\right)^{2a-1}$

If $a^m=a^n$ then $m=n$

$-3+8=2a-1\\2a=5+1\\a=\frac{6}{2}\\a=3$

b)

$\dfrac{6^{4} \times 9^{-2} \times 25}{2^{4} \times 5^{2} \times 3^{3}} = \dfrac{(2\times3)^{4} \times 9^{-2} \times 5^2}{2^{4} \times 5^{2} \times 3^{3}} \\=\dfrac{ 2^4\times 3^4\times 9^{-2} }{2^{4} \times 3^{3}}\\=\dfrac{ 3^4\times 9^{-2} }{3^{3}}$

Use the property of exponent, $\dfrac{a^m}{a^n}=a^{m-n}$

$=3^{4-3}\times 9^{-2}\\=3\times 9^{-2}$

Use the property of exponent, $a^{-n}=\dfrac{1}{a^n}$

$=3\times \dfrac{1}{9^2}\\=3\times \dfrac{1}{81}\\= \dfrac{1}{27}$