Let's go step by step through the questions in the image. I'll solve each question clearly, and underline each solution for clarity.

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5. Find integers x and y if $2x \times 3y = 6^4$

First, express $6^4$ as a product of primes:

$6^4 = (2 \times 3)^4 = 2^4 \times 3^4$

Now, compare this with the left-hand side:

$2^x \times 3^y = 2^4 \times 3^4$

This gives two equations:

$x = 4 \quad \text{and} \quad y = 4$

So, the solution is:

$x = 4, \quad y = 4$

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6. Find the value of $x$.

i. $\left(\dfrac{3^3 \times 36}{3^7 \times 3^5}\right) = 3^x$

Simplify the fraction:

$36 = 3^2 \times 2^2$

Thus, the equation becomes:

$\dfrac{3^3 \times 3^2 \times 2^2}{3^{7+5}} = 3^x$

This simplifies to:

$\dfrac{3^5 \times 2^2}{3^{12}} = 3^x$

So:

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

Since the powers of 3 must match, the solution is:

$x = -7$

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ii. $\left(\dfrac{7^x \times 7^3}{7^4 \times 7^2}\right)^2 = 7^3$

Combine powers on the left-hand side:

$\left(\dfrac{7^{x+3}}{7^6}\right)^2 = 7^3$

Simplifying gives:

$\left(7^{x-3}\right)^2 = 7^3$

This leads to:

$7^{2x - 6} = 7^3$

So:

$2x - 6 = 3$

Solving for $x$:

$2x = 9 \quad \Rightarrow \quad x = \dfrac{9}{2}$

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iii. $4^x = \dfrac{1}{64}$

Rewrite 64 as a power of 4:

$4^x = 4^{-3}$

Thus:

$x = -3$

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iv. $2^x = 0.125$

Rewrite 0.125 as a fraction:

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

Thus:

$x = -3$

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7. Given that $\dfrac{(36x^4)^2}{8x^2 \times 3x} = 2^a 3^b x^c$, find a, b, c.

First, simplify the left-hand side:

$(36x^4)^2 = 36^2 \times x^8 = (6^2)^2 \times x^8 = 6^4 \times x^8$

Now, simplify further:

$6^4 \times x^8 = (2^4 \times 3^4) \times x^8$

Now simplify the denominator:

$8x^2 \times 3x = 2^3 \times x^3 \times 3$

Now, divide the terms:

$\dfrac{2^4 \times 3^4 \times x^8}{2^3 \times 3^1 \times x^3} = 2^{4-3} \times 3^{4-1} \times x^{8-3}$

Simplifying:

$= 2^1 \times 3^3 \times x^5$

Thus, $a = 1$, $b = 3$, and $c = 5$.

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8. Given that $\dfrac{\sqrt{x^{-1} \times 3 \sqrt{y^2}}}{\sqrt{x^6 y^{2/3}}} = x^a y^b$, find a and b.

Simplify the expression inside the square root:

$\sqrt{x^{-1} \times 3y} = \sqrt{x^{-1} \times y}$

Now simplify the denominator:

$\sqrt{x^6 y^{2/3}} = x^3 y^{1/3}$

Now divide:

$\dfrac{\sqrt{x^{-1} y}}{x^3 y^{1/3}} = \dfrac{x^{-1/2} y^{1/2}}{x^3 y^{1/3}} = x^{-1/2 - 3} \times y^{1/2 - 1/3}$

Simplify exponents:

$= x^{-7/2} y^{1/6}$

Thus:

$a = -\dfrac{7}{2}, \quad b = \dfrac{1}{6}$

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I will continue with the remaining questions if you'd like me to proceed. Would you prefer that?