lilykhan566
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Sagot :

LammettHash

Note that each of the bases here are powers of 3:

3¹ = 3

3² = 9

3³ = 27

3⁴ = 81

3⁵ = 243

So the given equations can be rewritten as

[tex]\dfrac{81^{x+2y}}{27^{x-y}} = \dfrac{\left(3^4\right)^{x+2y}}{\left(3^3\right)^{x-y}} = \dfrac{3^{4x+8y}}{3^{3x-3y}} = 3^{x+11y} = 3^5[/tex]

and

[tex]\dfrac{243^{x+3y}}{81^{x+2y}} = \dfrac{\left(3^5\right)^{x+3y}}{\left(3^4\right)^{x+2y}} = \dfrac{3^{5x+15y}}{3^{4x+8y}} = 3^{x+7y} = 3^{-3}[/tex]

The bases on either side are the same, so the exponents must match and

x + 11y = 5

x + 7y = -3

Solve for x and y. Eliminate x and solve for y :

(x + 11y) - (x + 7y) = 5 - (-3)

(x - x) + (11y - 7y) = 5 + 3

4y = 8

y = 2

Solve for x :

x + 11•2 = 5

x + 22 = 5

x = -17

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