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2015-04-25T04:07:39+05:30

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Is the first question properly typed ?  is it y or x as the power of x ?

For the first question there are two possible sets of x and y.

2^x+3^y=17\\\\17=8+9=2^3+3^2,\ \ \ x=3,\ \ and\ \ y=2\\.\ \ =16+1=2^4+3^0,\ \ \ x=4\ and\ y=0


There are at least two solutions to the second question.  Perhaps there are more solutions.

2^{x+2}-3^{y-1}=5\\\\5=8-3=2^3-3^1,\\=\ \textgreater \  x+2=3,\ x=1,\ \ and\ \ y-1=1,\ \ y=2\\\\5=32-27=2^5-3^3\\=\ \textgreater \ x+2=5,\ \ x=3,\ \ \ and\ \ \ y-1=3,\ \ y=4

Perhaps there is a general formula possible to derive.
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Let
2^{a+1}-3^{b+1}=5=2+3\\2^{a+1}-2=3^{b+1}+3\\2(2^a-1)=3(3^b+1)\\\\So\ 2^a-1\ is\ multiple\ of\ 3\ and\ 3^b+1\ is\ a\ multiple\ of\ 2.

Another way Let:
2^{a+3}-3^{b+1}=5=2^3-3\\2^{a+3}-2^3=3^{b+1}-3\\2^3(2^a-1)=3(3^b-1)\\\\So\ 2^a-1\ is\ a\ multiple\ of\ 3.\ and\ 3^b-1\ is\ a\ multiple\ of\ 8.

we have to find a and b such that the above relations are  valid.

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