SOLUTION: If x^2+y^2=146xy then show that 2*log(x-y)=4*log2+2*log3+logx+logy

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Question 1199481: If x^2+y^2=146xy then show that 2*log(x-y)=4*log2+2*log3+logx+logy
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!

Let's start with the equation
2*log(x-y) = 4*log(2)+2*log(3)+log(x)+log(y)
and use log rules to say the following shown below.

2*log(x-y) = 4*log(2)+2*log(3)+log(x)+log(y)
log( (x-y)^2 ) = log(2^4)+log(3^2)+log(x)+log(y)
log( (x-y)^2 ) = log(16)+log(9)+log(x)+log(y)
log( (x-y)^2 ) = log(16*9*x*y)
log( (x-y)^2 ) = log(144xy)
(x-y)^2 = 144xy

Then expand out the left side and do a bit of moving terms around
(x-y)^2 = 144xy
x^2-2xy+y^2 = 144xy
x^2+y^2 = 144xy+2xy
x^2+y^2 = 146xy

We started with
2*log(x-y) = 4*log(2)+2*log(3)+log(x)+log(y)
and ended with
x^2+y^2 = 146xy

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That process can be done in reverse to go from
x^2+y^2 = 146xy
to
2*log(x-y) = 4*log(2)+2*log(3)+log(x)+log(y)

The steps would look like this
x^2+y^2 = 146xy
x^2+y^2 = 144xy+2xy
x^2-2xy+y^2 = 144xy
(x-y)^2 = 144xy
log( (x-y)^2 ) = log(144xy)
log( (x-y)^2 ) = log(16*9*x*y)
log( (x-y)^2 ) = log(16)+log(9)+log(x)+log(y)
log( (x-y)^2 ) = log(2^4)+log(3^2)+log(x)+log(y)
2*log(x-y) = 4*log(2)+2*log(3)+log(x)+log(y)

Answer by ikleyn(52794)   (Show Source):
You can put this solution on YOUR website!
.

                The problem formulation is FATALLY INCORRECT.

The correct formulation is THIS

        If x^2 + y^2 = 146xy   then show that   2*log(|x-y|) = 4*log(2)+2*log(3) + log(|x|) + log(|y|).

Using the absolute value signs is MANDATORY in this statement/identity.

Otherwise, you must restrict that everything under the logarithm function is positive.

It is because it is a rude mistake to state that

        log(x^2) = 2*log(x);   log(xy) = log(x) + log(y)

without making restrictions that x and y are positive.

Without such restriction, you only may state that

        log(x^2) = 2*log(|x|);   log(xy) = log(|x|)+log(|y|).

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