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Answer 1

Required Answer:-

Given:

• $\rm \big( \sqrt{27} \big)^{x} \div {3}^{y + 4 } = 1$
• $\rm {8}^{ \big(4 - \frac{x}{3} \big)} - {16}^{y} = 0$

To find:

• The values of x and y.

Solution:

For the first case,

$\rm \implies \big( \sqrt{27} \big)^{x} \div {3}^{y + 4 } = 1$

$\rm \implies \big( \sqrt{ {3}^{3} } \big)^{x} \div {3}^{y + 4 } = 1$

$\rm \implies \big( {3}\big)^{ \frac{3x}{2} } \div {3}^{y + 4 } = 1$

$\rm \implies \big( {3}\big)^{ \bigg(\dfrac{3x}{2} - y - 4 \bigg)} = 1$

$\rm \implies \big( {3}\big)^{ \bigg(\dfrac{3x}{2} - y - 4 \bigg)} = {3}^{0} \: \: \: \: (as \: {x}^{0} = 1)$

Comparing Base, we get,

$\rm \implies \dfrac{3x}{2} - y - 4 =0$

$\rm \implies 2 \times \bigg (\dfrac{3x}{2} - y - 4 \bigg) =2 \times 0$

$\rm \implies 3x - 2y - 8 =0$

$\rm \implies 3x - 2y = 8 \: \: ....(i)$

For the second case,

$\rm \implies {8}^{ \big(4 - \frac{x}{3} \big)} - {16}^{y} = 0$

$\rm \implies {8}^{ \big(4 - \frac{x}{3} \big)} = {16}^{y}$

$\rm \implies {2}^{3 \times \big(4 - \frac{x}{3} \big)} = {2}^{4y}$

$\rm \implies {2}^{(12 - x )} = {2}^{4y}$

Comparing Base, we get,

$\rm \implies 12 - x = 4y$

$\rm \implies x + 4y = 12 \: \: ....(ii)$

Multiplying both sides of equation (i) by 2, we get,

$\rm \implies 6x - 4y = 16 \: \: ....(iii)$

Adding equations (ii) and (iii), we get,

$\rm\implies 7x = 28$

$\rm\implies x = 4$

Now, substituting the value of x in equation (ii), we get,

$\rm \implies 4+ 4y = 12$

$\rm \implies 4y = 8$

$\rm \implies y = 2$

Hence, the values of x and y are 4 and 2 respectively.

Answer:

• The values of x and y are 4 and 2 respectively.

Answer 2

Answer:

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