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I have a question that I cannot solve. This mathematical problem has been taken from the book Mathematical Problem written by Faniye Alan and Ahmet Hançerlioğlu.

$$\log_x y =\log_y x $$ $$ x\neq y; x \neq 1; y \neq 1$$

what is x * y?

A)1/2 B)1 C)2 D)10 E)12

Base Change Rules of Logarithm

$$\log_x y=\frac{\log_z y}{\log_z x}$$ $$\log_x y=\frac{1}{\log_y x}$$

My Deduction $$\log_x y=\frac{1}{\log_y x} $$ $$ {\log_y x} =\log_x y $$ $$\log_x y=\frac{1}{\log_x y} $$ $$ (\log_x y)^2=1 $$

I cannot go further. How can I solve it?

Vivaan Daga
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tahasozgen
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  • The only possibility is $\log_xy=-1$. What does that tell you about the relationship between $x$ and $y$ – Rushabh Mehta Aug 02 '21 at 16:55
  • You have the identity that $\log_x y = \frac 1{\log_y x}$ so you have $\log_y x=\frac 1{\log_y x}$. How many numbers have the property $M = \frac 1M$? Well $M = \pm 1$ are the only option. If $\log_ x y = 1$ then $x^1 = y$ and $x = y$. That is not so so $\log_x y = -1$ and $x^{-1} = \frac 1x = y$. This can be any option: $\log_{2751} \frac {1}{2751} = -1 = \log_{\frac 1{2751}} 2751$ for example so no problem.... And so $xy = \frac 1y \cdot y = 1$. The answer is $xy = 1$. Note: no-one every asked you what $x$ or $y$ were, just what the product was. – fleablood Aug 02 '21 at 17:21
  • thank you fleablood – tahasozgen Aug 03 '21 at 11:53

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