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

froufroufox

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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.

IMG_1806.jpeg
 
Sorry, but I think you need to state the problem more clearly for someone to be able to help you. What does EE stand for in the equation above?
 
The equation you've written is

E=2x2+10y10xy5=2xy5+10y5xE=\frac{2x^2+10y^{10}}{xy^5}=\frac{2x}{y^5}+\frac{10y^5}{x}
By substitution of x=555x=5^{5^5} and y=554y=5^{5^4} into the expression, we obtain

2xy5+10y5x=25555545+105545555=2(575)+10(575)10(575)\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})
 
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The equation you've written is

E=2x2+10y10xy5=2xy5+10y5xE=\frac{2x^2+10y^{10}}{xy^5}=\frac{2x}{y^5}+\frac{10y^5}{x}
By substitution of x=555x=5^{5^5} and y=554y=5^{5^4} into the expression, we obtain

2xy5+10y5x=25555545+105545555=2(575)+10(575)10(575)\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})
Not quite; (554)5=(5(54))5=55(54)=555\left(5^{5^4}\right)^5=\left(5^{(5^4)}\right)^5=5^{5(5^4)}=5^{5^5}.

@froufroufox, can you take it from there?
 
Not quite; (554)5=(5(54))5=55(54)=555\left(5^{5^4}\right)^5=\left(5^{(5^4)}\right)^5=5^{5(5^4)}=5^{5^5}.

@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.
 
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