x y (3^x)(9^(42y))(81^(5x)) = (81^(525))(27^(21y)), In the table

Question

x > y

(3^x)(9^{42y})(81^{5x}) = (81^{525})(27^{21y})

In the table, select x and y such that the values are consistent with the information that is given. Make only two selections, one in each column.

scranjith · Answer

3^x . 3^84y . 3^20x = 3^4(525) 3^3(21y)
x+84y+20x = 4(525)+3(21y)
21(4y+x) = 21 (100+3y)
4y-3y+x = 100 => x+y = 100

X=73, Y=27

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Sajjad1994 · Answer

The OA of this question has been posted, It is:

x : 73
 y : 27

rocky620 · Answer

(3^x)(9^{42y})(81^{5x})=(81^{525})(27^{21y})

(3^x) = (3^x) 
 (9^{42y}) = (3^{84y}) 
 (81^{5x}) = (3^{20x})  
 (81^{525}) = (3^{2100}) 
 (27^{21y}) = (3^{63y})

So,

x + 84y + 20x = 2100 + 63y
21x + 21y = 2100
x+y = 100 .....EQ1

Only 27 and 73 satisfies EQ1. Since x>y

x=73
y=27

x > y (3^x)(9^(42y))(81^(5x)) = (81^(525))(27^(21y)), In the table