Question

If (x^-1y^2÷x^3y^-2)^1/3 ÷(x^6y^-3÷x^-2y^-3)^1/2 = x^a y^b, prove that a+b= -1, where x and y are different positive primes.

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{ { \bigg( \frac{ {x}^{ - 1} {y}^{2} }{ {x}^{3} {y}^{ - 2} } \bigg)}^{1}. { \bigg( \frac{ {x}^{ 6} {y}^{ - 3} }{ {x}^{2} {y}^{3} } \bigg)}^{ \frac{1}{2} } = {x}^{a} {y}^{b} }$

TO PROVE

a + b = -1, where x and y are different positive primes.

FORMULA TO BE IMPLEMENTED

We are aware of the formula on indices that :

$\sf{1. \: \: {a}^{m} \times {a}^{n} = {a}^{m + n} }$

$\displaystyle \sf{2. \: \: \: \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n} }$

$\displaystyle \sf{3. \: \: \: { ({a}^{m} )}^{n} = {a}^{mn} }$

$\displaystyle \sf{4. \: \: {a}^{0} = 1}$

EVALUATION

$\displaystyle \sf{ { \bigg( \frac{ {x}^{ - 1} {y}^{2} }{ {x}^{3} {y}^{ - 2} } \bigg)}^{1}. { \bigg( \frac{ {x}^{ 6} {y}^{ - 3} }{ {x}^{2} {y}^{3} } \bigg)}^{ \frac{1}{2} } = {x}^{a} {y}^{b} }$

$\displaystyle \sf{ \implies { \bigg( {x}^{ - 1 - 3} {y}^{2 + 2} \bigg)}^{1}. { \bigg( {x}^{ 6 - 2} {y}^{ - 3 - 3} }{ \bigg)}^{ \frac{1}{2} } = {x}^{a} {y}^{b} }$

$\displaystyle \sf{ \implies { \bigg( {x}^{ - 4} {y}^{4} \bigg)}^{1}. { \bigg( {x}^{ 4} {y}^{ - 6} }{ \bigg)}^{ \frac{1}{2} } = {x}^{a} {y}^{b} }$

$\displaystyle \sf{ \implies { \bigg( {x}^{ - 4} {y}^{4} \bigg)}^{1}. { \bigg( {x}^{ 2} {y}^{ - 3} }{ \bigg)}^{2 \times \frac{1}{2} } = {x}^{a} {y}^{b} }$

$\displaystyle \sf{ \implies { \bigg( {x}^{ - 4} {y}^{4} \bigg)}^{1}. { \bigg( {x}^{ 2} {y}^{ - 3} }{ \bigg)}^{1 } = {x}^{a} {y}^{b} }$

$\displaystyle \sf{ \implies { \bigg( {x}^{ - 4 + 2} {y}^{4 - 3} \bigg)} = {x}^{a} {y}^{b} }$

$\displaystyle \sf{ \implies {x}^{a} {y}^{b} = {x}^{ - 2} {y}^{1} }$

Comparing both sides we get

a = - 2 & b = 1

∴ a + b = - 2 + 1 = - 1

Hence proved

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Answer 2

Step-by-step explanation:

answer is proved a and b = 1

and x and y are diffrent positive number