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How do I solve these simultaneous equations?

$2 log_x y+2log_yx = 5$

$xy=8$

I've tried to convert the first formula to fraction form and continue from there, but I can't seem to get anywhere. I've tried to do

$x = 8/y$

and substitute to the first equation, but I still can't seem to solve this. How do I go about in solving these types of equations?

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Say $\log_x y=a$

Therefore $$a+\frac{1}{a}=\frac{5}{2}$$ $$2a^2-5a+2=0$$ $$(2a-1)(a-2)=0$$ $$a=2,\frac{1}{2}$$

Hence we have $$\log_x y=2,\frac{1}{2}$$

Now $$xy=8$$ $$1+\log_x y= \log_x 8$$ Hence $$\log_x 8-1=2$$ $$\log_x 8=3$$ $$\log_8 x=\frac{1}{3}$$ $$x=8^{\frac{1}{3}}=2$$

Similarly we also have $$\log_x 8-1=\frac{1}{2}$$ $$\log_x 8=\frac{3}{2}$$ $$x=4$$

The solutions hence follow.

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Use $\log_xy=\dfrac{\log y}{\log x}$ (See this)

to form a relation between $$\log x,\log y$$

and $$xy=5\implies\log x+\log y=\log5$$

Can you solve the two simultaneous equations?

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HINT :

We can write $$\log_y x=\frac{\log_xx}{\log_xy}=\frac{1}{\log_xy}$$ So, setting $t=\log_xy$ gives $$2t+2\cdot\frac 1t=5\Rightarrow 2t^2-5t+2=0$$

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  • $\begingroup$ For the sake of completeness, $y$ must not be $1$. Otherwise you would divide by zero. $\endgroup$ – Björn Friedrich Oct 31 '15 at 17:01

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