The system of logarithmic equations

Question

I have no idea how to solve the following system of equations: $$\begin{cases}\log_{|xy|}{(x-y)}=1\\ 2\log_{5}{|xy|}\log_{|xy|}{(x+y)}=1\end{cases}$$

Answer 1

HINT

Note that for $|xy|\neq0,1$

$$2\log_{5}{|xy|}\log_{|xy|}{(x+y)}=1\iff \frac{\log_{|xy|}{(x+y)^2}}{\log_{|xy|}{5}}=1\iff \log_{|xy|}{(x+y)^2}=\log_{|xy|}{5}$$

$$\implies (x+y)^2=5$$

and

$$\log_{|xy|}{(x-y)}=1 \implies x-y = |xy|$$

thus

• $(x+y)^2=5\implies x+y=\sqrt 5$ since x+y>0 for existence of $\log$
• $x-y = |xy|$ consider 2 cases

1) $xy>0$

$\implies x-y=xy\implies x +(x-\sqrt 5)=x(\sqrt 5-x) \implies x^2+(2-\sqrt 5)x-\sqrt 5=0$

$\implies x_1=\frac{\sqrt 5 +1}{2}\quad x_2=\frac{\sqrt 5-5}{2} \implies y_1=\frac{\sqrt 5 -1}{2}\quad y_2=\frac{\sqrt 5+5}{2}$

2) $xy<0$

$\implies x-y=-xy\implies x +(x-\sqrt 5)=-x(\sqrt 5-x) \implies x^2-(2+\sqrt 5)x+\sqrt 5=0$

$\implies x_1=\frac{\sqrt 5 -1}{2}\quad x_2=\frac{5+\sqrt 5}{2} \implies y_1=\frac{\sqrt 5 +1}{2}\quad y_2=\frac{-5+\sqrt 5}{2}$

Answer 2

Hint: $$\log_5(|xy|) = \frac{1}{\log_{|xy|}(5)}.$$ Consequently, $$2\log_{5}{|xy|}\log_{|xy|}{(x+y)}=1 \implies x+y =\sqrt{5}.$$ And $$\log_{|xy|}{(x-y)}=1 \implies x-y = |xy|.$$

Answer 3

$$1=2\cdot\log_{5}{|xy|}\log_{|xy|}{(x+y)}$$

Using change of base on $\log_5|xy|\implies\dfrac{1}{\log_{|xy|}5}$

$$1=2\cdot\dfrac{1}{\log_{|xy|}5}\cdot\log_{|xy|}{(x+y)}\implies 1=2\cdot\dfrac{\log_{|xy|}(x+y)}{\log_{|xy|}5}\implies1=2\cdot\log_5(x+y)$$

Note that $\dfrac{\log_{|xy|}(x+y)}{\log_{|xy|}5}=\log_5(x+y)$ because of change of base.