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Bingel [31]
3 years ago
14

If (1/4)^(x+y) = 256 and log₄(x-y) = 3, what are the exact values of x and y

Mathematics
1 answer:
andrew-mc [135]3 years ago
6 0

Answer:

x = 30 and y = -34

Step-by-step explanation:

Given the following functions

(1/4)^(x+y) = 256... 1

log₄(x-y) = 3.... 2

From equation 2;

x-y = 4³

x-y = 64

x = 64 + y ... 3

Substitutw 3 into 1

From 1:

(1/4)^(x+y) = 256

(1/4)^(64+y+y) = 256

(1/4)^(64+2y) = 256

Take log₄ of both sides

64+2y log₄ (1/4) = log₄256

-(64+2y) = 4log₄4

-(64+2y) = 4

64+2y = -4

2y = -4 - 64

2y = -68

y = -34

Since

x = 64 + y .

x = 64 - 34

x = 30

Hence x = 30 and y = -34

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Let the number be (10 x + y)

Then, 10 x + y = 3 * (x + y) (given)

Or, 10 x + y = 3 x + 3 y

Or, 10 x - 3 x = 3 y - y

Or, 7x = 2 y

Or, x = 2 y / 7 ( Eq. 1)

Also, 10 x + y + 45 = 10 y + x ( given)

Or, 10 x - x = 10 y - y - 45

Or, 9 x = 9 y - 45 ( Eq. 2)

Substituting the value of x from (Eq. 1) in (Eq. 2), we have:

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Answer

Check:

Sum of the digits = 2 + 7 = 9

9 * 3 = 27 ✓

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