Logarithmic sums

Question

If $x^2+y^2-3xy=0$, and $x>y$, then find $\log (x-y)$ base $xy$. Please help me out. I am trying it but can't do it.. help me. How to do the sum? I tried solving the equation first but the answer is not coming

Answer 1

From $x^2+y^2-3xy=0$, you get that $x^2+y^2-2xy=xy$; in other words, $(x-y)^2=xy$ and therefore $\log\bigl((x-y)^2\bigr)=\log(xy)$. Another way of stating this is $2\log(x-y)=\log(xy)$. So$$\log_{xy}(x-y)=\frac{\log(x-y)}{\log(xy)}=\frac12.$$

Answer 2

$(x-y)^2 = xy$, hence $xy \gt 0$ , since $x \gt y. $

Take $\log_{xy}$ of both sides:

$2 \log_{xy}(x-y) = \log_{xy} (xy) = 1.$