Given $x^2 + y^2 = 34xy$, show that $\log\left(\frac{x+y}{6}\right)= \frac{\log x + \log y}{2}$

Question

If $x^2 + y^2 = 34xy$, show that $$\log\left(\frac{x+y}6\right)= \frac{\log x + \log y}{2}.$$

I tried to put log into the first equation, but I have no idea about how the $34$ being simplified in the second equation.

Can anybody show me the full work solution?

Answer 1

From $x^2 +y^2 = 34xy \implies x^2 + 2xy + y^2 = 36xy$ by adding $2xy$ to both sides we get $$(x+y)^2 = 36xy \iff (x+y)^2 = 6^2xy$$

Now dividing by $6^2$ and making use of the fact that $a^2/b^2 = (a/b)^2$ we get $$\left(\frac{x+y}{6}\right)^2 = xy$$

Now taking the logarithm of both sides yields $$\log \left(\frac{x+y}{6}\right)^2 = \log xy$$

Bringing the $2$ in the power down and using the product rule for logarithms yields $$2 \log \left(\frac{x+y}{6}\right) = \log x + \log y$$

Now dividing by $2$ gives us $$\bbox[10px, border: blue solid 1px]{\log \left(\frac{x+y}{6}\right) = \frac{\log x + \log y}{2}}$$

Answer 2

$$ 2\lg\left(\frac{x+y}{6}\right) = \lg x + \lg y\Leftrightarrow \lg\left(\frac{x+y}{6}\right)^2 = \lg(xy) \Leftrightarrow (x+y)^2=36xy $$ I suppose you can proceed.

Answer 3

Multiply both sides by 2:

$$2\log\left(\frac{x+y}{6}\right) = \log x + \log y$$

$$\text{LHS} = 2\log\left(\frac{x+y}{6}\right) = \log\left(\frac{x+y}{6}\right)^2$$

$$\text{LHS} = \log\left(\frac{x^2 +y^2+2xy}{36}\right)$$

replace $x^2 + y^2$ with $34xy$

$$ \text{LHS} = \log\left(\frac{34xy+2xy}{36}\right) = \log\left(\frac{36xy}{36}\right) $$

Can you finish it from here? You need to show that $$\log\left(\frac{36xy}{36}\right) = \log x + \log y $$

Answer 4

For $x\gt 0,y\gt 0$, we have$$\begin{align}x^2+y^2=34xy&\Rightarrow x^2+2xy+y^2=34xy+2xy\\&\Rightarrow (x+y)^2=36xy\\&\Rightarrow \left(\frac{x+y}{6}\right)^2=xy\\&\Rightarrow 2\log\left(\frac{x+y}{6}\right)=\log(xy)\\&\Rightarrow \log\left(\frac{x+y}{6}\right)=\frac 12(\log(x)+\log(y))\end{align}$$