I have a few expressions and need to get a final simplified (reduced,solved) results. I am new and tried/searched different ways for my problem but couldn’t find the reason. These are my expressions:
R0 = FractionalPart[Log[5, x + (3/ 4)]] == FractionalPart[Log[5, xP + (3/ 4)]]
R1 = Log[5, xP + (3 / 4)] - Log[3, yP + ( 5 / 2)] == Log[5, x + (3 / 4)] - Log[3, y + ( 5 / 2)]
F = ( (x == xP && y == yP) || ( Exists[{xPP, yPP}, xPP < yPP && xP == 5 xPP + 3 && yP == 3 yPP + 5]))
R = R0 && R1 && F
Reduce[R && xP >= yP , {x, y, xP, yP}, Reals]
when I run this it takes a long time and seems infinite loop or something and never finishes. I also tried this
Reduce[R , Element[{x, y, xP, yP}, Reals]]
and it returns the error: Reduce : This system cannot be solved with the methods available to Reduce
We expect two functions for xP and yP based on x and y, for each. Please help me through this.
2- How can I make sure that each single expression is error free?
Thanks a lot
and that is close enough to zero that this might be an acceptable solution for you.
You must check all this very carefully to try to make certain that I haven’t made any mistakes.
If I start from a different position and see from the statement of your problem that x==xP and y==yP then it seems that R0, R1 and F are always true for any values of x and y. Thus the only condition seems to be that x>y. Is that correct?
Solve[FractionalPart[Log[5, x + (3/4)]] ==
FractionalPart[Log[5, X + (3/4)]] &&
Log[5, x + (3/4)] - Log[3, y + (5/2)] ==
Log[5, X + (3/4)] - Log[3, Y + (5/2)] &&
((x == X && y == Y)) &&
X >= Y,
{X, Y}, Integers]
Here is one solution of the other subsystem:
In[24]:= FindInstance[
FractionalPart[Log[5, x + (3/4)]] ==
FractionalPart[Log[5, X + (3/4)]] &&
Log[5, x + (3/4)] - Log[3, y + (5/2)] ==
Log[5, X + (3/4)] -
Log[3, Y + (5/2)] && (x < y && (X - 3)/5 < (Y - 5)/3) && X >= Y &&
Element[X | Y, Integers], {X, Y, x, y}]
Out[24]= {{X -> 13, Y -> 12, x -> 2, y -> 7/3}}