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

Question

Алгебра

Morluntrius

5 год назад

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

ОТВЕТЫ

Гермоген

Jun 30, 2019

$9^{x} - 2^{x-0.5} = 2^{x+2.5} - 3^{2x-1} amp;#10;\\\amp;#10; 9^{x} +3^{2x-1} = 2^{x-0.5} + 2^{x+2.5} amp;#10;\\\amp;#10;9^x+ \dfrac{3^{2x}}{3} = \dfrac{2^x}{2^{0.5}} +2^x\cdot 2^{2.5}amp;#10;\\\\amp;#10;9^x+ \dfrac{9^x}{3} = \dfrac{2^x}{2^{0.5}} +2^x\cdot 2^{2.5}amp;#10;\\\\amp;#10;9^x+ \dfrac{1}{3} \cdot 9^x= \dfrac{\sqrt{2}}{ 2 } \cdot 2^x+4 \sqrt{2}\cdot 2^xamp;#10;\\\amp;#10;(1+ \dfrac{1}{3}) \cdot 9^x= (\dfrac{\sqrt{2}}{ 2 } +4 \sqrt{2})\cdot 2^x$
$\dfrac{4}{3} \cdot 9^x= \dfrac{9\sqrt{2}}{ 2 } \cdot 2^xamp;#10;\\\amp;#10; \dfrac{9^x}{2^x} = \dfrac{9\sqrt{2}}{ 2 } : \dfrac{4}{3}amp;#10;\\\amp;#10;\left( \dfrac{9}{2} \right)^x= \dfrac{27\sqrt{2}}{ 8 } amp;#10;\\\amp;#10;\left( \dfrac{9}{2} \right)^x= \dfrac{27}{ 4 \sqrt{2} } amp;#10;\\\amp;#10;x=\log_{4.5}\frac{27}{ 4 \sqrt{2} } =\log_{4.5}(\frac{9}{ 2 }\cdot \frac{9}{ 2 }\cdot \frac{1}{3 \sqrt{2} } )=2+\log_{4.5}\frac{1}{3 \sqrt{2} } =2-\log_{4.5}3 \sqrt{2}$
Ответ: $2-\log_{4.5}3 \sqrt{2}$

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