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

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You need to solve for x and y the system of simultaneous equations, such that:

`{(log_4(x + y) = 2),(log_3 x + log_3 y = 2 + log_3 7):}`

You need to move all terms that contain logarithms, in the bottom equation, to the left side, such that:

`log_3 x + log_3 y - log_3 7 = 2`

Using logarithmic identities yields:

`log_3 (x*y)/7 = 2 => (x*y)/7 = 3^2 => (x*y)/7 = 9 => (x*y) = 63`

Converting the top logarithmic equation into an exponential equation yields:

`log_4(x + y) = 2 => x + y = 4^2 => x + y = 16`

You may form the quadratic equation such that:

`x^2 - 16x + 63 = 0 => x^2 - 16x = -63`

Completing the square yields:

`x^2 - 16x + 64 = 64 - 63`

`(x - 8)^2 = 1 => x - 8 = +-1 => x_1 = 9; x_2 = 7`

**Hence, evaluating the solutions to the given simultaneous equations yields `x = 9, y = 7` or `x = 7, y = 9.` **

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}