Complicated Logarithm

Question

If $x>0$, $y>0$, and

$$x^2 + y^2=98xy$$

then $\log(x+y)$ can be expressed as $A\log(x)+B\log(y)+C$ where $A,B,C$ are real numbers and all logarithms are base $10$ logarithms. Compute $100ABC$.

Answer 1

Here's a hint: Add $2xy$ to both sides of your equation. Then recall that $x^2+2xy+y^2=(x+y)^2$.

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$$\begin{align*}x^2+2xy+y^2&=100xy\\ (x+y)^2&=100xy\\ 2\log(x+y)&=\log(100)+\log(x)+\log(y)\\ 2\log(x+y)&=\log(x)+\log(y)+2\\ \log(x+y)&=\underbrace{\frac{1}{2}}_{A}\log(x)+\underbrace{\frac{1}{2}}_{B}\log(y)+\underbrace{1_\strut}_{C}\\\\ 100ABC&=100\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot 1=25 \end{align*}$$