Change each of the bases to the simpliest base.

9 = 3²

9^x = 3^{2 x x} = 3^2x

729 = 3⁶

$\frac{1}{729}$ = $\frac{1}{3^6}$

But a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

$\frac{1}{3^6}$ = 3^-6

If 9^x x 3^2y = $\frac{1}{729}$

3^2x x 3^2y = 3^-6

Note:

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

3^{2x + 2y} = 3^-6

Since the bases are the same, the powers are also the same.

2x + 2y = -6

Divide both sides by 2 to simplify further.

x + y = - 3 --- eqn 1

2^-x x 4^-y = $\frac{1}{8}$

4 = 2²

8 = 2³

$\frac{1}{8}$ = $\frac{1}{2^3}$ = 2^-3

2^-x x 4^-y = $\frac{1}{8}$

2^-x x 2^{2 x -y} = 2^-3

2^-x x 2^-2y = 2^-3

2^{-x + -2y} = 2^-3

Since the bases are the same, the powers are also the same.

-x - 2y = - 3

Divide both sides by -1

x + 2y = 3

x = 3 - 2y --- eqn 2

Substitute --- eqn 2 into --- eqn 1

3 - 2y + y = -3

3 - y = -3

3 + 3 = y

6 = y

y = 6

Substitute y = 6 into --- eqn 2

x = 3 - 2 x 6

x = 3 - 12

x = -9

$\sqrt[3]{x}$ = ${\mathrm{x}}^{\frac{1}{3}}$

$\sqrt[3]{2^2}$ = $2^{\mathrm{2\; x}\frac{1}{3}}$ = $2^{\frac{2}{3}}$

Change the base 16 to base 2 so that both the left and right sides have the same bases.

16 = 2 x 2 x 2 x 2 = 2⁴

16ⁿ = $\sqrt[3]{2^2}$, find the value of n.

2^{4 x n} = $2^{\frac{2}{3}}$

2⁴ⁿ = $2^{\frac{2}{3}}$

Since the bases are the same, the powers are also the same.

4n = $\frac{2}{3}$

Get rid of the fraction by multiplying both sides by the denominator, 3

3 x 4n = $\frac{2}{3}$ x 3

12n = 2

Divide both sides by 12.

n = $\frac{2}{12}$ = $\frac{1}{6}$