Laws of Exponents Problem: x = 5^5^5, y = 5^5^4, E = (2x^2 + 10y^{10})/(xy^5)

[froufroufox]

Hey, I hope you can help me with this one.
I know the answer is 12 but the idea is to solve it using laws of exponents.

I guess i need to do something like ((2+10)(2x^2)(10y^10))/xy^5 and then reduce things with an exponent to 1 so it goes like (2+10) (1), maybe?, but I haven’t been able to get there.

 

[Aion]

Sorry, but I think you need to state the problem more clearly for someone to be able to help you. What does [imath]E[/imath] stand for in the equation above?

 

[Aion]

The equation you've written is

[math]E=\frac{2x^2+10y^{10}}{xy^5}=\frac{2x}{y^5}+\frac{10y^5}{x}[/math]
By substitution of [imath]x=5^{5^5}[/imath] and [imath]y=5^{5^4}[/imath] into the expression, we obtain

[math]\frac{2x}{y^5}+\frac{10y^5}{x}=2\frac{5^{5^5}}{5^{5^{4^5}}}+10\frac{5^{5^{4^5}}}{5^{5^{5}}}=2(5^{-75})+10(5^{75}) \approx10(5^{75})[/math]

 

[Dr.Peterson]

The equation you've written is

[math]E=\frac{2x^2+10y^{10}}{xy^5}=\frac{2x}{y^5}+\frac{10y^5}{x}[/math]
By substitution of [imath]x=5^{5^5}[/imath] and [imath]y=5^{5^4}[/imath] into the expression, we obtain

[math]\frac{2x}{y^5}+\frac{10y^5}{x}=2\frac{5^{5^5}}{5^{5^{4^5}}}+10\frac{5^{5^{4^5}}}{5^{5^{5}}}=2(5^{-75})+10(5^{75}) \approx10(5^{75})[/math]

Not quite; [imath]\left(5^{5^4}\right)^5=\left(5^{(5^4)}\right)^5=5^{5(5^4)}=5^{5^5}[/imath].

@froufroufox, can you take it from there?

 

[Aion]

Not quite; [imath]\left(5^{5^4}\right)^5=\left(5^{(5^4)}\right)^5=5^{5(5^4)}=5^{5^5}[/imath].

@froufroufox, can you take it from there?

I was going to bed when you pointed out my silly mistake. Good thing I didn't solve it completely though.