اگر ${\log ^{\left( {{x^۲} - ۱} \right)}} = {\log ^{\left( {x - ۱} \right)}} + ۲{\log ^۳}$ آنگاه $\log _۲^x$ کدام است؟
${\log ^{\left( {{x^2} - 1} \right)}} - {\log ^{\left( {x - 1} \right)}} = 2{\log ^3}$$ \Rightarrow {\log ^{\frac{{{x^2} - 1}}{{x - 1}}}} = {\log ^{{3^2}}} \Rightarrow \frac{{{x^2} - 1}}{{x - 1}} = 9$$ \Rightarrow \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right)}} = 9$$x + 1 = 9 \Rightarrow \,\,\,\,\,\,\,\,\,\,,\log _2^x = \log _2^8 = \log _2^{{2^3}}$$ = 3\log _2^2 = 3 \times 1 = 3$