Evaluate 9⁻¹ x ((3)/(4))2 and solve for x in 16^x = 8^x+1 and 813x x 272x = 729.

$$\frac{1}{9} \times \frac{9}{16} = \frac{1 \times 9}{9 \times 16} = \frac{9}{144} = \frac{1}{16}$$

The evaluated expression is $\boxed{\frac{1}{16}}$.

b) Using indices find the value of $x$

a) $16^x = 8^{x+1}$

$$16 = 2^4$$ $$8 = 2^3$$

So the equation becomes:

$$(2^4)^x = (2^3)^{x+1}$$

$$2^{4x} = 2^{3(x+1)}$$ $$2^{4x} = 2^{3x+3}$$

$$4x - 3x = 3$$ $$x = 3$$

The value of $x$ is $\boxed{3}$.

b) $81^{3x} \times 27^{2x} = 729$

$$81 = 3^4$$ $$27 = 3^3$$ $$729 = 3^6$$

So the equation becomes:

$$(3^4)^{3x} \times (3^3)^{2x} = 3^6$$

$$3^{4 \times 3x} \times 3^{3 \times 2x} = 3^6$$ $$3^{12x} \times 3^{6x} = 3^6$$

$$3^{12x+6x} = 3^6$$ $$3^{18x} = 3^6$$

$$x = \frac{6}{18}$$ $$x = \frac{1}{3}$$

The value of $x$ is $\boxed{\frac{1}{3}}$.

c) $\frac{64 \times 128}{8} = 2^{2x-4}$

$$64 = 2^6$$ $$128 = 2^7$$ $$8 = 2^3$$

So the equation becomes:

$$\frac{2^6 \times 2^7}{2^3} = 2^{2x-4}$$

First, apply the product rule $a^m \times a^n = a^{m+n}$ in the numerator:

$$\frac{2^{6+7}}{2^3} = 2^{2x-4}$$ $$\frac{2^{13}}{2^3} = 2^{2x-4}$$

Next, apply the quotient rule $\frac{a^m}{a^n} = a^{m-n}$:

$$2^{13-3} = 2^{2x-4}$$ $$2^{10} = 2^{2x-4}$$

$$10 + 4 = 2x$$ $$14 = 2x$$ $$x = \frac{14}{2}$$ $$x = 7$$

The value of $x$ is $\boxed{7}$.