# What is $\log_{a}{x} \cdot \log_{y}{a}$ given below system of equations? 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: $$\log_{a}{x} \cdot \log_{a}{y}= ?$$

## 5 Answers

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 $$\log_ax\log_ay=1.$$

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.

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.

## Hint

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

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

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?