# Find x and y: log 4 (x+y) = 2 log 3 x + log 3 y = 2 + log 3 7

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### 2 Answers

We'll impose the constraints of existence of logarithms:

x>0

y>0

Now, we'll take anti-logarithm for the first equation:

x+y = 4^2

x+y = 16 (1)

We'll apply the product rule of logarithms in the second equation:

log 3 x + log 3 y = log 3 (x*y)

We'll re-write the sum from the right side of the second equation:

2 + log 3 7 =2*1 + log 3 7

2*1 + log 3 7 = 2*log 3 3 + log 3 7

We'll apply the power rule of logarithms:

2*log 3 3 = log 3 3^2 = log 3 9

We'll re-write the second equation:

log 3 (x*y) = log 3 9 + log 3 7

log 3 (x*y) = log 3 (9*7)

log 3 (x*y) = log 3 (63)

Since the bases are matching, we'll apply one to one rule:

x*y = 63 (2)

We'll write x with respect to y, from (1) and we'll substitute in (2):

x = 16 - y

(16 - y)*y = 63

We'll remove the brackets:

16y - y^2 - 63 = 0

We'll re-arrange the terms and we'll multiply by -1:

y^2 - 16y + 63 = 0

We'll apply the quadratic formula:

y1 = [16+sqrt(256 - 252)]/2

y1 = (16+2)/2

y1 = 9

y2 = 7

x1 = 16 - y1

x1 = 16-9

x1 = 7

x2 = 16-7

x2 = 9

**The solutions of the symmetric system are:**

**{9 ; 7} and {7 ; 9}**

log4(x+y) = 2... (1).

log3 x +log 3y = 2+log3 7.....(2).

We solve for x and y:

We know if log a(b) = x, then b = a^x.

Therefore from (1) we get: x+y = 4^2 = 16.

x+y = 16. ...(3).

From (2) log 3 (x*y) = log 3 (9 )+log3(7) , as log a + log b = logab and 2 = log3(9) .

log3 (xy) = log3 (9*7) .

We take anti log of both sides:

xy = 63....(4).

Therefore from (3) and (4), (x-y)^2 = x+y)^2 - 4xy = 16^2 -63*4 = 4.

Therefore x- y = sqrt4 = 2.

Therefore x-y = 2 ...(5). and x+y = 16...(3).

(3)+(5) gives: 2x= 2+16 = 18, So 2x= 18, x = 18/2 = 9.

(3)-(5) gives: 2y = 16-2 = 14. So x= 14/2 = 7.

Therefore x = 9 and y = 7 are the solutions.