Find number of solutions to the equation 3 x − 1 + 5 x − 1 = 3 4 .
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Let n = x - 1. By Fermat's Last Theorem, there are no integer solutions when n is bigger than 2. Of course, for n=2, we have the well-known Pythagorean triples. Hence, when x = 3, 3^2 + 5^2 = 34. For all other known values of n (1, 0, -1, and -2), the equation does not satisfy 3^n + 5^n = 34. The only value to satisfy the equation is n = 2, or x = 3.
How did we know that it's a bijection function?
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It is continuous for x ∈ ( − ∞ , ∞ ) this means there is a f ( x ) for every x . Since f ( x ) is strictly increasing there is only one value of f ( x ) for every x .
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Let f ( x ) = 3 x − 1 + 5 x − 1 . Note that f ( x ) has a range of ( 0 , ∞ ) for x ∈ ( − ∞ , ∞ ) . Note also that f ′ ( x ) = ln 3 ⋅ 3 x − 1 + ln 5 ⋅ 5 x − 1 > 0 for all real x . This means that f ( x ) is a strictly increasing function and it is a one-to-one corresponding or bijection function, there is only one value of f ( x ) or a value of x for all real x . Therefore, there is only 1 solution for f ( x ) = 3 x − 1 + 5 x − 1 = 3 4 . The solution is x = 3 .