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Solve a system equation 
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#1
Dec1812, 03:05 PM

P: 17

Hi everybody,
Please I want to Solve the system: 3^{x}2^{y}=19 y^{3}2^{x}=19 x,y real number! Thank you ! 


#2
Dec1812, 05:07 PM

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P: 12,045

I think clever guessing (and showing that there are not other solutions) is the quickest method. You can solve one of those equations for a single variable and put this into the other equation, but I don't think you cannot simply find solutions like that.



#3
Dec1812, 05:35 PM

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P: 11,948

It looks like 3,3 is the only solution.



#4
Dec1912, 06:46 AM

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P: 12,045

Solve a system equation
There is another solution near y=20.8216, x=13.1370.



#5
Dec1912, 09:02 AM

P: 17

Please I want the steps to find it, because I'm stumbled in the last steps as follow:
x*Ln(3)y*Ln(2)=Ln(19) 3*Ln(y)x*Ln(2)=Ln(19) x*Ln(3)y*Ln(2)  3*Ln(y)x*Ln(2) = 0 x*Ln(3)+x*Ln(2) = y*Ln(2)+3*Ln(y) {each one equals Ln(19)} x*Ln(3)+x*Ln(2) = y*Ln(2)+3*Ln(y) = Ln(19) x*Ln(6) = y*Ln(2)+3*Ln(y) = Ln(19) x = ln(19)/ln(6) ..... Here I'm stumbled. Thank you for your response. 


#6
Dec1912, 09:12 AM

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P: 12,045

There is no useful way to simplify ln(19)/ln(6). It is just some number  and it is wrong.
The first step does not work. ln(ab) is not the same as ln(a)  ln(b). The third step looks wrong, too (where you say "{each one equals Ln(19)}"). 


#7
Dec1912, 11:20 AM

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P: 21,397

You apparently took the natural log of both sides, like so: ln(3^{x}  2^{y}) = ln(19) Then you "distributed" the ln operation like this: ln(3^{x})  ln(2^{y}) = ln(19) This is the step that is incorrect. There is no property of logs in which ln(A + B) = ln(A) + ln(B). 


#8
Dec1912, 02:41 PM

P: 17

Thank you all.



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