adsense
How many integers are there? \(x\) satisfy \({\rm{lo}}{{\rm{g}}_3}\frac{{{x^2} – 16}}{{343}} < {\rm{lo}}{{\rm{ g}}_7}\frac{{{x^2} – 16}}{{27}}\)?
A. 193.
B. 92.
C. 186.
D. 184.
The answer:
adsense
Choose EASY
TXĐ: \(D = \left( { – \infty ; – 4} \right) \cup \left( {4; + \infty } \right).\)
We have:
\(\begin{array}{l}{\rm{lo}}{{\rm{g}}_3}\frac{{{x^2} – 16}}{{343}} < {\rm{ lo}}{{\rm{g}}_7}\frac{{{x^2} – 16}}{{27}}\\ \Leftrightarrow {\rm{lo}}{{\rm{g}} _3}7.\left[ {{\rm{lo}}{{\rm{g}}_7}\left( {{x^2} – 16} \right) – 3} \right] < {\rm{lo}}{{\rm{g}}_7}\left( {{x^2} – 16} \right) – 3{\rm{lo}}{{\rm{g}} _7}3\\ \Leftrightarrow \left( {{\rm{lo}}{{\rm{g}}_3}7 – 1} \right){\rm{.lo}}{{\rm{g} }_7}\left( {{x^2} – 16} \right) < 3{\rm{lo}}{{\rm{g}}_3}7 – 3{\rm{lo}}{{\ rm{g}}_7}3\\ \Leftrightarrow {\log _7}\left( {{x^2} – 16} \right) < \frac{{3\left( {{{\log }_3}7 – {{\log }_7}3} \right)}}{{{{\log }_3}7 – 1}}\\ \Leftrightarrow {\log _7}\left( {{x^2} – 16} \right) < 3\left( {1 + {{\log }_7}3} \right)\\ \Leftrightarrow {\log _7}\left( {{x^2} – 16} \right) < {\ log _7}{21^3}\\ \Leftrightarrow {x^2} – 16 < {21^3}\\ \Leftrightarrow – \sqrt {9277} < x < \sqrt {9277} \end{array}\)
Combining the conditions we have \(x \in \left\{ { – 96; – 95;…; – 5;5;…;95;96} \right\}\). So there are 184 integers x that satisfy.
