Question

[Phan Thị Lê Anh]

So sánh :

a) \(\log_32\) và \(\log_23\)

b) \(\log_23\) và \(\log_311\)

c) \(\frac{1}{2}+lg3\) và \(lg19-lg2\)

a) \(lg\frac{5+\sqrt{7}}{2}\) và \(\frac{lg5+lg\sqrt{7}}{2}\)  

Answer 1

[Đỗ Đức Huy]

a) Ta có \(\log_32<\log_33=1=\log_22<\log_23\)

b) \(\log_23<\log_24=2=\log_39<\log_311\)

c) Đưa về cùng 1 lôgarit cơ số 10, ta có 

\(\frac{1}{2}+lg3=\frac{1}{2}lg10+lg3=lg3\sqrt{10}\)

\(lg19-lg2=lg\frac{19}{2}\)

So sánh 2 số \(3\sqrt{10}\) và \(\frac{19}{2}\) ta có :

\(\left(3\sqrt{10}\right)^2=9.10=90=\frac{360}{4}<\frac{361}{4}=\left(\frac{19}{2}\right)^2\)

Vì vậy : \(3\sqrt{10}<\frac{19}{2}\)

Từ đó suy ra \(\frac{1}{2}+lg3\)<\(lg19-lg2\)

d) Ta có : \(\frac{lg5+lg\sqrt{7}}{2}=lg\left(5\sqrt{7}\right)^{\frac{1}{2}}=lg\sqrt{5\sqrt{7}}\)

Ta so sánh 2 số : \(\sqrt{5\sqrt{7}}\) và \(\frac{5+\sqrt{7}}{2}\) 

Ta có :

\(\sqrt{5\sqrt{7}}^2=5\sqrt{7}\)

\(\left(\frac{5+\sqrt{7}}{2}\right)^2=\frac{32+10\sqrt{7}}{4}=8+\frac{5}{2}\sqrt{7}\)

\(8+\frac{5}{2}\sqrt{7}-5\sqrt{7}=8-\frac{5}{2}\sqrt{7}=\frac{16-5\sqrt{7}}{2}=\frac{\sqrt{256}-\sqrt{175}}{2}>0\)

Suy ra : \(8+\frac{5}{2}\sqrt{7}>5\sqrt{7}\)

Do đó : \(\frac{5+\sqrt{7}}{2}>\sqrt{5\sqrt{7}}\)

và \(lg\frac{5+\sqrt{7}}{2}>\frac{lg5+lg\sqrt{7}}{2}\)