Please solve for x, 64^x+1=16^2y+5

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1 Answer

I read this as: 64^(x+1)=16^(2y+5).

64=4³ and 16=4², 4^(3x+3)=4^(4y+10), so 3x+3=4y+10, 3x-4y-7=0. This is an equation in two variables and has many solutions for x and y. However, if the question is wrongly presented and y should be x, so 3x-4x-7=0, -x-7=0, so x=-7.

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