Question

Розв'яжіть рівняння:
$4 \times 2 {}^{2x} - 6 {}^{x} = 18 \times 3 {}^{2x}$
​

Answer 1

Ответ:

$x=-2$

Объяснение:

$4 \times 2^{2x} - 6^{x} = 18 \times 3^{2x}\\\\4 \times 2^{2x} - 6^{x} -18 \times 3^{2x}=0\\\\4\times 2^{2x}-2^x\times3^x-18\times3^{2x}=0 \qquad |\div3^{2x}$

$4\times\dfrac{2^{2x}}{3^{2x}}-2^x\times3^{x}\div 3^{2x}-18\times\dfrac{3^{2x}}{3^{2x}}=0\\\\\\4\times\left(\dfrac{2}{3}\right)^{2x}-2^x\times3^{x-2x}-18=0\\\\\\4\times\left(\dfrac{2}{3}\right)^{2x}-2^x\times3^{-x}-18=0\\\\\\4\times\left(\dfrac{2}{3}\right)^{2x}-2^x\times\dfrac{1}{3^x}-18=0\\\\\\4\times\left(\dfrac{2}{3}\right)^{2x}-\dfrac{2^x}{3^x}-18=0\\\\\\4\times\left(\dfrac{2}{3}\right)^{2x}-\left( \dfrac{2}{3}\right)^x-18=0$

$4\times\left(\left(\dfrac{2}{3}\right)^x\right)^{2}-\left( \dfrac{2}{3}\right)^x-18=0\\\\\\\bold{t=\left(\dfrac{2}{3}\right)^x}\\\\\downarrow\\\\4t^2-t-18=0\\\\4t^2+8t-9t-18=0\\\\4t(t+2)-9(t+2)=0\\\\(4t-9)(t+2)=0\\\\4t_1-9=0 \qquad \qquad t_2+2=0\\\\4t_1=9 \qquad \qquad \quad \ \ \boxed{t_2=-2}\\\\\boxed{t_1=\frac{9}{4}}$

$\left( \dfrac{2}{3}\right)^{x}=\dfrac{9}{4} \qquad \qquad \left( \dfrac{2}{3}\right)^{x}=-2$

                                     $\downarrow$

$\left( \dfrac{2}{3}\right)^{x}=\left(\dfrac{3}{2}\right)^2 \qquad \quad x \notin \mathbb{R}\\\\\\\left( \dfrac{2}{3}\right)^{x}=\left(\dfrac{2}{3}\right)^{-2}$

   $\downarrow$

$x=-2$

$\downarrow$    $\downarrow$    $\downarrow$

$\boxed{x=-2}$