Idea solve a system of equations

Question

can you give me the idea to solve this system, $$ \begin{equation} \left\lbrace \begin{array}{} \log_y(x)+\log_x(y)=\frac{5}{2}\\ xy=27 \end{array} \right. \end{equation}$$

Answer 1

Using that $$\log_{x}{y}=\frac{1}{\log_{x}{y}}$$ we get $$1+\frac{1}{(\log_{x}{y})}=\frac{5}{2}\log_{x}{y}$$ substituting $t=\log_{x}{y}$ you have a quadratic equation. $$1+t=\frac{5}{2}t^2$$

Answer 2

$\log_x y$ and $\log_y x$ are reciprocals of each other, so you have $$ u + \frac 1 u = \frac 5 2. $$ Multiplying both sides by $u$ yields $$ u^2 + 1 = \frac 5 2 u $$ or, if you like, $$ 2u^2 + 2 = 5u. $$ So $\log_x y = 2$ and $\log_y x = \frac 1 2;$ and so $x^2 = y.$

Now you have \begin{align} & x^2 = y, \\ & xy = 27. \end{align} From this you get $x^3 = 27.$

Answer 3

A possible way is as follows:

• We can write $\log_x y = \frac{\log_b y}{\log_b x}$ with any base $b>0, b\neq 1$.
• Since $xy=27=3^3$, you may choose $b= 3$.
• Besides this you can take the logarithm of the second equation. So, you get the system $$\log_3 (xy) = \log_3 x + \log_3 y = 3, \frac{\log_3 x}{\log_3 y} + \frac{\log_3 y}{\log_3 x} = \frac{5}{2}$$

Now, just substitute $\log_3 y= 3 - \log_3 x$ in the other equation and solve for $\log_3 x$. After some rearranging you get

$$\log_3^2 x - 3 \log_3 x + 2 = 0 \Leftrightarrow (\log_3 x -1)(\log_3 x -2) = 0$$

Since $\log_3 x =1 \Leftrightarrow x=3$ and $\log_3 x =2 \Leftrightarrow x=9$, the solutions are $x=3, y=9$ and $x=9,y=3$.