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(52800) (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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The lowest possible score to the " problem's creator ",
a ticket and a serious warning.
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