Phương trình đã cho tương đương với:
$(5-\sqrt{21})^x+7(5+\sqrt{21})^x=8.2^x$
$\Leftrightarrow \left(\dfrac{5-\sqrt{21}}{2}\right)^x+7\left(\dfrac{5+\sqrt{21}}{2} \right)^x=8 (1)$
Ta có:$\left(\dfrac{5-\sqrt{21}}{2}\right)^x\left(\dfrac{5+\sqrt{21}}{2} \right)^x=1 \Rightarrow \left(\dfrac{5-\sqrt{21}}{2}\right)^x=\dfrac{1}{\left(\dfrac{5+\sqrt{21}}{2} \right)^x}$
Đặt $t=\left(\dfrac{5+\sqrt{21}}{2}\right)^x\Rightarrow\left(\dfrac{5-\sqrt{21}}{2} \right)^x=\dfrac{1}{t}$. Khi đó $(1)$ trở thành:
$\dfrac{1}{t}+7t=8$
$\Leftrightarrow 7t^2-8t+1=0$
$\Leftrightarrow \left[\begin{array}{l}t=1\\t=\dfrac{1}{7}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left(\dfrac{5+\sqrt{21}}{2}\right)^x=1\\\left(\dfrac{5+\sqrt{21}}{2}\right)^x=\dfrac{1}{7}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = 0\\\left(\dfrac{5-\sqrt{21}}{2}\right)^x= 7\end{array} \right.$
$\Leftrightarrow \left[\begin{array}{l}x = 0\\x=\log_{\left(\frac{5-\sqrt{21}}{2}\right)}7\end{array}\right.$