Question

2×81^x+1 - 36^x+1 - 3×16^x+1 = 0

Answer 1

$2 cdot 81^{x + 1} - 36^{x + 1} - 3 cdot 16^{x + 1} = 0 \ \ 2 cdot 9^{2(x + 1)} - 9 ^{x + 1} cdot 4^{x + 1} - 3 cdot 4^{2(x + 1)} = 0 |:4^{2(x + 4) }\ \ 2 cdot (dfrac{9}{4} )^{2(x + 1)}- (dfrac{9}{4} )^{x + 1} - 3 = 0 \ \ t = (dfrac{9}{4} )^{x + 1}, t textgreater 0 \ \ 2t^2 - t - 3 = 0 \ \ D = 1 + 3 cdot 4 cdot 2 = 25 = 5^2 \ \ t_1 = dfrac{1 + 5}{4} = dfrac{3}{2}\ \ t_2 = dfrac{1 - 5}{4}= -4 - ne ed.$

Обратная замена:
$(dfrac{9}{4} )^{x + 1} = dfrac{3}{2} \ \ (dfrac{3}{2} )^{2x + 2} = ( dfrac{3}{2} )^{1} \ \ 2x + 2 = 1 \ \ 2x = 1 - 2 \ \ x = -0,5$