Let x=3^a. The given equation is equivalent to x+x^2+x^3-84=0. Since 84=4+16+64, x=4 is an obvious solution. By polynomial division this factors as (x-4)(x^2+5x+21). From the second factor, by application of the formula for quadratic equations we have two complex solutions x=(-5±i√(59))/2. Since a=log(x)/log(3), the "Nice Olympiad Math Exponential Problem" has three solutions a=log(4)/log(3) and a=log((-5±i√(59))/2)/log(3).
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@@Easy_maths540 Sen iyi bir dahisin
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Let x=3^a. The given equation is equivalent to x+x^2+x^3-84=0. Since 84=4+16+64, x=4 is an obvious solution. By polynomial division this factors as (x-4)(x^2+5x+21). From the second factor, by application of the formula for quadratic equations we have two complex solutions x=(-5±i√(59))/2. Since a=log(x)/log(3), the "Nice Olympiad Math Exponential Problem" has three solutions a=log(4)/log(3) and a=log((-5±i√(59))/2)/log(3).
Espectacular!
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