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Điều kiện $x>1.$ PT
$\Leftrightarrow (x- 1) ^{\log_2 8(x-1)} = 8.(x-1)^3$
$\Leftrightarrow (x- 1) ^{\log_2 \left[ {(x-1)} \right]+3} = 8.(x-1)^3$
$\Leftrightarrow (x- 1)^3 .(x- 1) ^{\log_2 (x-1)} = 8.(x-1)^3$
$\Leftrightarrow \left[ {\begin{matrix} x=1\\(x- 1) ^{\log_2 (x-1)} = 8 \end{matrix}} \right.$
$\Leftrightarrow \left[ {\begin{matrix} x=1\\\log_2(x- 1) ^{\log_2 (x-1)}= \log_28 \end{matrix}} \right.$
$\Leftrightarrow \left[ {\begin{matrix} x=1\\\log_2(x- 1).\log_2 (x-1)= 3 \end{matrix}} \right.$
$\Leftrightarrow \left[ {\begin{matrix} x=1\\\log_2 (x-1)= \pm \sqrt 3 \end{matrix}} \right.$
$\Leftrightarrow \left[ {\begin{matrix} x=1\\x= 2^{\pm \sqrt 3}+1 \end{matrix}} \right.$
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