## 1. a) Show that the number $\log _{2015} 2016$ is irrational;

## b) Compare the numbers $\log _{5} 6$ and $\log _{6} 7$;

c) Calculate $E=\lg ^{3} 5+\lg ^{3} 20+\lg 8 \cdot \lg (0.25)$.
Solution: a) Suppose the number $\log _{2015} 2016$ is rational. Then there exist numbers $a, b \in N^{*}$ such that $\log _{2015} 2016=\frac{a}{b}$. We obtain $2016= 2015^{\frac{a}{b}} \Leftrightarrow 2016^{b}=2015^{a}$, a relation impossible since one number is even and the other is odd. (2p)

b) Applying the inequality of means for the positive numbers $\log _{6} 5$ and $\log _{6} 7$ we get $\sqrt{\log _{6} 5 \cdot \log _{6} 7} \leq \frac{\log _{6} 5+\log _{6} 7}{2}=\frac{\log _{6} 35}{2}<\frac{\log _{6} 36}{2}=1$, thus $\log _{6} 5 \cdot \log _{6} 7<1 \Leftrightarrow \log _{6} 7<\log _{5} 6$. (2p)
c) $\quad E=\lg 35+(2 \lg 2+\lg 5)^{3}+3 \lg 2 \cdot(-2 \lg 2)=(1-\lg 2)^{3}+$ $(1+\lg 2)^{3}-6 \lg ^{2} 2=2(3p)$