25~26高三上·河南多校天一小高考·第8题

C

【题目】已知$5^a = 6$，若$m = 4^a – 5$，$n = 6^a – 7$，则（ ）
A．$m > n > 0$
B．$n > m > 0$
C．$n > 0 > m$
D．$m > 0 > n$

【答案】C

【解析】因为$5^a = 6$，则$a = \log_5 6 = \frac{\ln 6}{\ln 5}$，

由$\frac{\ln 6}{\ln 5} – \frac{\ln 6 + \ln \frac{4}{5}}{\ln 5 + \ln \frac{4}{5}} $$= \frac{\ln 6}{\ln 5} – \frac{\ln 6 + \ln 4 – \ln 5}{\ln 4} $$= \frac{\ln 4\ln 6 – \ln 5(\ln 6 + \ln 4 – \ln 5)}{\ln 4\ln 5} $$= \frac{(\ln 4 – \ln 5)(\ln 6 – \ln 5)}{\ln 4\ln 5} < 0$，

所以$a < \frac{\ln 6 + \ln \frac{4}{5}}{\ln 5 + \ln \frac{4}{5}} = \frac{\ln 6 + \ln 4 – \ln 5}{\ln 4} $$= \frac{\ln \frac{24}{5}}{\ln 4} < \frac{\ln 5}{\ln 4} = \log_4 5$，则$4^a < 5$．

由$\frac{\ln 6}{\ln 5} – \frac{\ln 6 + \ln \frac{7}{6}}{\ln 5 + \ln \frac{7}{6}} $$= \frac{\ln 6}{\ln 5} – \frac{\ln 7}{\ln 5 + \ln 7 – \ln 6} $$= \frac{\ln 6(\ln 5 + \ln 7 – \ln 6) – \ln 5\ln 7}{\ln 5(\ln 5 + \ln 7 – \ln 6)} $$= \frac{(\ln 5 – \ln 6)(\ln 6 – \ln 7)}{\ln 5(\ln 5 + \ln 7 – \ln 6)} > 0$

所以$a > \frac{\ln 6 + \ln \frac{7}{6}}{\ln 5 + \ln \frac{7}{6}} = \frac{\ln 7}{\ln 5 + \ln 7 – \ln 6} $$= \frac{\ln 7}{\ln \frac{35}{6}} > \frac{\ln 7}{\ln 6} = \log_6 7$，所以$6^a > 7$，

所以$m = 4^a – 5 < 0 < n = 6^a – 7$，则$n > 0 > m$．

故选：C.