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a)\[3^x=2^xy+1\]
b)\[x^y=y^{x-y}\]

3^x = 2^x y + 1
y =0, x=0 <-- this works ... since the lhs is power of 3

x=1, y=1 also works where you will get 3 = 3

so we just try out numbers

if 3^n -1 is of form divisible by 2^n, then we have solutions of it.

actually i never knew that they are called that diophantine

\[3^x-2^xy=1\]
but we can try x=1,2,3,4
and see if they yield the y value to give 1

for these types of question, you should probably ask mukushla
till then let me fool around.

lol thanks a lot

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thanks man!! you always amaze me!!

np bro :)

haha