An algebra problem by Saswata Naha

Algebra Level 1

Find number of solutions to the equation 3 x 1 + 5 x 1 = 34 3^{x-1}+5^{x-1}=34 .


The answer is 1.

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2 solutions

Let f ( x ) = 3 x 1 + 5 x 1 f(x) = 3^{x-1}+5^{x-1} . Note that f ( x ) f(x) has a range of ( 0 , ) (0, \infty) for x ( , ) x \in (-\infty, \infty) . Note also that f ( x ) = ln 3 3 x 1 + ln 5 5 x 1 > 0 f'(x) = \ln 3\cdot 3^{x-1} + \ln 5 \cdot 5^{x-1} > 0 for all real x x . This means that f ( x ) 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 ) f(x) or a value of x x for all real x x . Therefore, there is only 1 \boxed{1} solution for f ( x ) = 3 x 1 + 5 x 1 = 34 f(x) = 3^{x-1}+5^{x-1}= 34 . The solution is x = 3 x=3 .

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.

Mariano Gomez Bent - 4 years ago

How did we know that it's a bijection function?

Kaushik Chandra - 3 years, 12 months ago

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It is continuous for x ( , ) x \in (-\infty, \infty) this means there is a f ( x ) f(x) for every x x . Since f ( x ) f(x) is strictly increasing there is only one value of f ( x ) f(x) for every x x .

Chew-Seong Cheong - 3 years, 11 months ago
Razing Thunder
Jul 3, 2020
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a=0
for x in range(1,100):
    if 3**(x-1) + 5**(x-1)==34:
        a+=1
print(a)        

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