a) Áp dụng công thức: \(\log_ab.\log_bc=\log_ac\)

b) Vì \(\dfrac{1}{\log_{a^k}b}=\dfrac{1}{\dfrac{1}{k}\log_ab}=\dfrac{k}{\log_ab}\) nên biểu thức vế trái bằng:

\(VT=\dfrac{1}{\log_ab}\left(1+2+...+n\right)\)

\(=\dfrac{1}{\log_ab}.\dfrac{n\left(n+1\right)}{2}=VP\)

a.

ĐKXĐ: \(x>0\)

\(log_5x>6\Rightarrow x>6^5\Rightarrow x>7776\)

b.

ĐKXĐ: \(x>0\)

\(log_7x< 2\Rightarrow\left\{{}\begin{matrix}x>0\\x< 7^2\end{matrix}\right.\) \(\Rightarrow0< x< 49\)

c. 

\(log_2x\le3\Rightarrow\left\{{}\begin{matrix}x>0\\x\le3^2\end{matrix}\right.\) \(\Rightarrow0< x\le9\)

d.

\(log_{\dfrac{1}{3}}x>27\Rightarrow\left\{{}\begin{matrix}x>0\\x< \left(\dfrac{1}{3}\right)^{27}\end{matrix}\right.\)

\(\Rightarrow0< x< \dfrac{1}{3^{27}}\)

a. Vì \(0< 0,1< 1\) nên bất phương trình đã cho 

\(\Leftrightarrow0< x^2+x-2< x+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x-2>0\\x^2-5< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< -2\\x>1\end{matrix}\right.\\-\sqrt{5}< x< \sqrt{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{5}< x< -2\\1< x< \sqrt{5}\end{matrix}\right.\)

Vậy tập nghiệm của bất phương trình là \(S=\left\{-\sqrt{5};-2\right\}\) và \(\left\{1;\sqrt{5}\right\}\)

b. Điều kiện \(\left\{{}\begin{matrix}2-x>0\\x^2-6x+5>0\end{matrix}\right.\)

Ta có:

 \(log_{\dfrac{1}{3}}\left(x^2-6x+5\right)+2log^3\left(2-x\right)\ge0\)

\(\Leftrightarrow log_{\dfrac{1}{3}}\left(x^2-6x+5\right)\ge log_{\dfrac{1}{3}}\left(2-x\right)^2\)

\(\Leftrightarrow x^2-6x+5\le\left(2-x\right)^2\)

\(\Leftrightarrow2x-1\ge0\)

Bất phương trình tương đương với:

\(\left\{{}\begin{matrix}x^2-6x+5>0\\2-x>0\\2x-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>5\end{matrix}\right.\\x< 2\\x\ge\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{1}{2}\le x< 1\)

Vậy tập nghiệm của bất phương trình là: \(\left(\dfrac{1}{2};1\right)\)

a) \(log_3\dfrac{6}{5}>log_3\dfrac{5}{6}\) vì \(\dfrac{6}{5}>\dfrac{5}{6}\)

b) \(log_{\dfrac{1}{3}}9>log_{\dfrac{1}{3}}17\) vì \(9>17\) và \(0< \dfrac{1}{3}< 1\).

c) \(log_{\dfrac{1}{2}}e>log_{\dfrac{1}{2}}\pi\) vì \(e>\pi\) và \(0< \dfrac{1}{2}< 1\)

d) \(log_2\dfrac{\sqrt{5}}{2}>log_2\dfrac{\sqrt{3}}{2}\)  vì \(\dfrac{\sqrt{5}}{2}>\dfrac{\sqrt{3}}{2}\).

ĐKXĐ: \(x>\dfrac{1}{2}\)

\(log_{\dfrac{1}{2}}\left(\dfrac{x+1}{2x-1}\right)< 2\)

\(\Rightarrow\dfrac{x+1}{2x-1}>\dfrac{1}{4}\)

\(\Rightarrow x>-\dfrac{5}{2}\)

Kết hợp ĐKXĐ: \(\Rightarrow x>\dfrac{1}{2}\)

1.\(\dfrac{log_ac}{log_{ab}c}=log_ac.log_c\left(ab\right)=log_ac.\left(log_ca+log_cb\right)=log_ac.log_ca+log_ac.log_cb=\dfrac{log_ac}{log_ac}+\dfrac{log_cb}{log_ca}=1+log_ab\)

2. \(log_{ax}bx=\dfrac{log_abx}{log_aax}=\dfrac{log_ab+log_ax}{log_aa+log_ax}=\dfrac{log_ab+log_ax}{1+log_ax}\)

3. \(\dfrac{1}{log_ax}+\dfrac{1}{log_{a^2}x}+...+\dfrac{1}{log_{a^n}x}=log_xa+log_xa^2+...+log_xa^n\)

\(=log_xa+2log_xa+...+n.log_xa=log_xa+2log_xa+...+n.log_xa\)

\(=log_xa.\left(1+2+...+n\right)=\dfrac{n\left(n+1\right)}{2}log_xa=\dfrac{n\left(n+1\right)}{2.log_ax}\)

1.

\(A=3log_{2^2}\sqrt{a}-log_{2^{-1}}a^2+2log_{a^{\dfrac{1}{2}}}a\)

\(=3.\dfrac{1}{2}.\dfrac{1}{2}log_2a-\left(-1\right).2.log_2a+2.2.log_2a\)

\(=\dfrac{27}{4}log_2a\)

2.

\(log_{12}36=\dfrac{log_236}{log_212}=\dfrac{log_2\left(3^2.2^2\right)}{log_2\left(3.2^2\right)}=\dfrac{log_23^2+log_22^2}{log_23+log_22^2}\)

\(=\dfrac{2.log_23+2}{log_23+2}=\dfrac{2a+2}{a+2}\)