Can You Solve These Unique Algebraic Equations?

[naoufelabs]

Hi everybody,
Please I want to Solve the system:

3^x-2^y=19
y³-2^x=19

x,y real number!

Thank you !

 

[mfb]

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.

 

[dextercioby]

It looks like 3,3 is the only solution.

 

[mfb]

There is another solution near y=20.8216, x=13.1370.

 

[naoufelabs]

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.

 

[mfb]

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(a-b) is not the same as ln(a) - ln(b).
The third step looks wrong, too (where you say "{each one equals Ln(19)}").

 

[Mark44]

3^x-2^y=19
y³-2^x=19
x,y real number!

The first equation below is wrong, so everything below it is also invalid.

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).

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.

 

[naoufelabs]

Thank you all.