# What is the ratio x/y whether 2log(2x-3y)=logx+logy ?

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

It is given that 2*log (2x - 3y) = log x + log y. We need to determine x/y.

2*log (2x - 3y) = log x + log y

Use the property that log a + log b = log a*b

=> 2*log (2x - 3y) = log x*y

use the property a*log b = log b^a

=> log (2x - 3y)^2 = log x*y

=> (2x - 3y)^2 = x*y

=> 4x^2 + 9y^2 - 12xy = xy

=> 4x^2 + 9y^2 - 13xy = 0

Divide by xy

=> 4x/y + 9y/x - 13 = 0

=> 4t + 9/t - 13 = 0

Solving for t we get the value of x/y

=> 4t^2 + 9 - 13t = 0

=> 4t^2 -4t - 9t + 9 = 0

=> 4t(t - 1) -9(t - 1) = 0

=> (4t - 9)(t - 1) = 0

=> t = 9/4 and t = 1

=> x/y = 1 and x/y = 9/4 = 2.25

The ratio x/y = 1 and x/y = 2.25 if 2*log (2x - 3y) = log x + log y

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll apply the power rule of logarithms for the term from the left side:

2*log (2x-3y) = log (2x-3y)^2

Since the bases of logarithms from the sum from the right side are matching, we'll apply the product rule:

logx+logy = log (x*y)

We'll re-write the identity:

log (2x-3y)^2 = log (x*y)

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

(2x-3y)^2 = x*y

We'll expand the square form the left:

4x^2 - 12x*y + 9y^2 - x*y = 0

4x^2 - 13x*y + 9y^2 = 0

We'll divide by y^2:

4(x/y)^2 - 13(x/y) + 9 = 0

We'll replace the fraction x/y by t:

4t^2 - 13t + 9 = 0

t1 = [13+sqrt(169 - 144)]/8

t1 = (13+5)/8

t1 = 18/8

t1 = 9/4

t2 = (13-5)/8

t2 = 1

The possible values of the ratio x/y are: x/y = 9/4 and x/y = 1.