Question
Find all real numbers x satisfying the equation:
2^x + 3^x - 4^x + 6^x - 9^x = 1.
Please help me. I really don't know what to do.
2^x + 3^x - 4^x + 6^x - 9^x = 1.
Please help me. I really don't know what to do.
Answers
Hmm. It's clear that x=0 works, but is there anything else?
(2^x+1)(3^x+1) = 2^x + 3^x + 6^x + 1
so,
(2^x+1)(3^x+1) +1-4^x + 1-9^x = 4
(1+2^x)(1+3^x) + (1+2^x)(1-2^x) + (1+3^x)(1-3^x) = 4
(1+2^x)(1+3^x+1-2^x) + (1+3^x)(1-3^x) = 4
I don't see anything dropping out. If we can show that 1 is the maximum value of the function, then there is probably no other solution. Can't see right off how to show that, either. There's probably some clever trick involved.
(2^x+1)(3^x+1) = 2^x + 3^x + 6^x + 1
so,
(2^x+1)(3^x+1) +1-4^x + 1-9^x = 4
(1+2^x)(1+3^x) + (1+2^x)(1-2^x) + (1+3^x)(1-3^x) = 4
(1+2^x)(1+3^x+1-2^x) + (1+3^x)(1-3^x) = 4
I don't see anything dropping out. If we can show that 1 is the maximum value of the function, then there is probably no other solution. Can't see right off how to show that, either. There's probably some clever trick involved.
Thanks, I already know that 0 is the answer but I don't know how to arrive to that. Huhu, I don't know the solution. Please help me, thanks.
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