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I let $\log_{a}{x}=m$ and $\log_{y}{a}=n$. So I have to find $m\cdot n$. From the system of equations we get

$$m-\frac{1}{n}=1 \quad \quad n-\frac{1}{m}=1$$

From here I find that $m=n$ (Consequently, $\log_{a}{x}=\log_{y}{a}$).

I can't progress any further from here. How can I solve this problem?

Edit: Most probably there has been a typo in the writing of the problem. The question should have been: $$ \log_{a}{x} \cdot \log_{a}{y}= ? $$

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We have $$\log_ax-\log_ay=1$$ and $$\frac{1}{\log_ax}-\frac{1}{\log_ay}=-1,$$ which gives $$\frac{-1}{\log_ax\log_ay}=-1$$ or $$\log_ax\log_ay=1.$$

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Plug $m=n$ in one of your equations and find $mn$. I think none of the options are right! The question should be $\log_ax\log_a y$ which gives 1 as the answer.

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Let's review how you got $m=n$. I expect you equated two expressions for $1$ to obtain $m-1/n=n-1/m$, then rearranged this as $m-n=(m-n)/mn$. Thus either $m=n$ (which leads us to the surds you mentioned in a comment, viz. $mn=m^2=m+1$) or $mn=1$, viz. D. Of course, that's nonsense because we'd then have $m-1/n=0$ instead. I expect whoever invented the problem overlooked this (what's its source?), or thought a trick such as $\infty-\infty=1$ would get around it. Edit: or they meant to ask for $\log_a x\log_a y=m/n$, as others have suggested.

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Hint

$$\log_y\frac{1}{a}.\log_{\frac{1}{a}}y=1$$

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  • $\begingroup$ I'm getting $\frac{3+\sqrt{5}}{2}$ & $\frac{3-\sqrt{5}}{2}$. I multiplied the two equations to use this property. $\endgroup$ – Eldar Rahimli Apr 11 at 6:27
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Hint:

$$\begin{aligned}\log_ax+\log_{1/a}y=1\iff\log_a\left(\dfrac{x}{y}\right)=1&\implies x=ay\\ \log_{x}a-\log_ya=-1=\dfrac{1}{\log_ax}-\dfrac{1}{\log_ay}=\dfrac{-1}{\log_ax\log_ay}&\implies \log_ax\log_ay=1\end{aligned}$$

Now put in $ay$ for $x$ to get the following: $$1=\left[1+\log_ay\right]\log_ay$$

Can you proceed?

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