$2^y = 5, 5^x = 2$,$xy = ?$

Question

$$2^y = 5$$ $$ 5^x = 2$$ $$xy = ?$$

Indeed, I could find the $x$ and $y$ by using logarithm. However, I don't want to do it with that way. Let me show my thinkings:

$$5^x \cdot 2^y = 10$$

Here we get

$$xy = 10$$

It seems wrong on my textbook, why?

My Kindest Regards!

Answer 1

Here is an approach without using logs: $$ 5^{xy} = \left(5^x\right)^y = 2^y = 5 $$ so $xy=1$.

Answer 2

Hint: $\;5 = 2^y = (5^x)^y = 5^{xy},$.

Answer 3

Don't feel bad about taking $ln$ because it is simple and also does not need the calculator you mentioned:

$2^y = 5 \to \ln(2^y)=\log(5) \to y\ln(2)=\ln(5) \to y=\dfrac{\ln(5)}{\ln(2)}$

$5^x = 2 \to \ln(5^x)=\ln(2) \to x\ln(5)=\ln(2) \to x=\dfrac{\ln(2)}{\ln(5)}$

$x \cdot y =\dfrac{\ln(5)}{\ln(2)} \cdot \dfrac{\ln(2)}{\ln(5)}=1$

Answer 4

$$ 2^x = 5 = 2^{1\over y}$$ so $x ={1\over y}$...

Answer 5

Given that $ 2^{y} = 5 $, $$ 5^{x} = (2^{y})^{x} = 2^{yx} = 2^{xy} = 2$$

Solving $ 2^{xy} = 2 $ for $xy$: $$ \log(2^{xy}) = xy\log(2) = \log(2)$$ or $$ xy =1$$