Question

Решите уравнение:
$9^{x} - 2^{x-0.5} = 2^{x+2.5} - 3^{2x-1}$

Answer 1

$9^{x} - 2^{x-0.5} = 2^{x+2.5} - 3^{2x-1} \ 9^{x} +3^{2x-1} = 2^{x-0.5} + 2^{x+2.5} \ 9^x+ dfrac{3^{2x}}{3} = dfrac{2^x}{2^{0.5}} +2^xcdot 2^{2.5} \\ 9^x+ dfrac{9^x}{3} = dfrac{2^x}{2^{0.5}} +2^xcdot 2^{2.5} \\ 9^x+ dfrac{1}{3} cdot 9^x= dfrac{sqrt{2}}{ 2 } cdot 2^x+4 sqrt{2}cdot 2^x \ (1+ dfrac{1}{3}) cdot 9^x= (dfrac{sqrt{2}}{ 2 } +4 sqrt{2})cdot 2^x$
$dfrac{4}{3} cdot 9^x= dfrac{9sqrt{2}}{ 2 } cdot 2^x \ dfrac{9^x}{2^x} = dfrac{9sqrt{2}}{ 2 } : dfrac{4}{3} \ left( dfrac{9}{2} right)^x= dfrac{27sqrt{2}}{ 8 } \ left( dfrac{9}{2} right)^x= dfrac{27}{ 4 sqrt{2} } \ x=log_{4.5}frac{27}{ 4 sqrt{2} } =log_{4.5}(frac{9}{ 2 }cdot frac{9}{ 2 }cdot frac{1}{3 sqrt{2} } )=2+log_{4.5}frac{1}{3 sqrt{2} } =2-log_{4.5}3 sqrt{2}$
Ответ: $2-log_{4.5}3 sqrt{2}$