Ta có \(a=\log_{25}7=\frac{\log_27}{\log_225}=\frac{\log_27}{2\log_25}=\frac{\log_27}{2b}\Rightarrow\log_27=2ab\)

\(\Rightarrow H=\log_{\sqrt[3]{5}}\frac{49}{8}=\frac{\log_2\frac{49}{8}}{\log_2\sqrt[3]{5}}=\frac{\log_2\frac{7^2}{2^2}}{\log_25^{\frac{1}{3}}}=\frac{2\log_27-3}{\frac{1}{3}\log_25}=\frac{12ab-9}{b}\)

\(log_{a^4}b^4.log_ba^5=\dfrac{1}{4}.4.log_ab.5.log_ba=5.log_ab.log_ba=5\)

\(log_{a^3}b^2.log_ba^4=\dfrac{1}{3}.2.log_ab.4.log_ba=\dfrac{8}{3}.log_ab.log_ba=\dfrac{8}{3}\)

\(log_{a^{15}}b^7.log_{b^{49}}a^{30}=\dfrac{1}{15}.7.log_ab.\dfrac{1}{49}.30.log_ba=\dfrac{2}{7}log_ab.log_ba=\dfrac{2}{7}\)

\(log_{a^{2021}}b^{2020}.log_{b^{4040}}a^{6063}=\dfrac{1}{2021}.2020.log_ab.\dfrac{1}{4040}.6063.log_ba=\dfrac{3}{2}\)

Ta có : \(\left(a^{\log_37}\right)^{\log_37}+\left(b^{\log_711}\right)^{\log_711}+\left(c^{\log_{11}25}\right)^{\log_{11}25}=27^{^{\log_37}}+49^{^{\log_711}}+\left(\sqrt{11}\right)^{^{\log_{11}25}}\)

                                                                                         \(=7^3+11^2+25^{\frac{1}{2}}=469\)

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.

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

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

a) \(A=\log_{5^{-2}}5^{\frac{5}{4}}=-\frac{1}{2}.\frac{5}{4}.\log_55=-\frac{5}{8}\)

b) \(B=9^{\frac{1}{2}\log_22-2\log_{27}3}=3^{\log_32-\frac{3}{4}\log_33}=\frac{2}{3^{\frac{3}{4}}}=\frac{2}{3\sqrt[3]{3}}\)

c) \(C=\log_3\log_29=\log_3\log_22^3=\log_33=1\)

d) Ta có \(D=\log_{\frac{1}{3}}6^2-\log_{\frac{1}{3}}400^{\frac{1}{2}}+\log_{\frac{1}{3}}\left(\sqrt[3]{45}\right)\)

                   \(=\log_{\frac{1}{3}}36-\log_{\frac{1}{3}}20+\log_{\frac{1}{3}}45\)

                   \(=\log_{\frac{1}{3}}\frac{36.45}{20}=\log_{3^{-1}}81=-\log_33^4=-4\)

a)\(\log_{\frac{2}{x}}x^2-14\log_{16x}x^3+40\log_{4x}\sqrt{x}=0\)ĐKXĐ: x>0

\(\Leftrightarrow2\log_{\frac{2}{x}}x-42\log_{16x}+20\log_{4x}\sqrt{x}=0\)

\(\Leftrightarrow\frac{2}{\log_x\frac{2}{x}}-\frac{42}{\log_x16x}+\frac{20}{\log_x4x}=0\)

\(\Leftrightarrow\frac{2}{\log_x2-1}-\frac{42}{4\log_x2+1}+\frac{20}{2\log_x+1}=0\)

Đặt \(\log_x2=a\left(a\in R\right)\)

Thay vào pt:\(\frac{2}{a-1}-\frac{42}{4a+1}+\frac{20}{2a+1}=0\)

\(\Leftrightarrow2a^2-a+4=0\)(pt này vô nghiệm)

Vậy pt đã cho vô nghiệm

cái đó phải là \(-42\log_{16x}x\) nhé bạn

\(\log_{\frac{x}{2}}4x^2+2\log_{\frac{x^3}{8}}2x+\log_{2x}\frac{x^4}{4}=-\frac{14}{3}\)(ĐKXĐ:x>0)

\(\Leftrightarrow2\log_{\frac{x}{2}}2x+\frac{2}{3}\log_{\frac{x}{2}}2x+2\log_{2x}\frac{x^2}{2}=-\frac{14}{3}\)

\(\Leftrightarrow\frac{8}{3}\log_{\frac{x}{2}}2x+2\log_{2x}\frac{x^2}{2}=-\frac{14}{3}\)

Xét \(\log_{2x}\frac{x^2}{2}=\log_{2x}\frac{x^2}{4}\cdot2=2\log_{2x}\frac{x}{2}+\log_{2x}2=\frac{2}{\log_{\frac{x}{2}}2x}+\frac{1}{1+\log_2x}\)

Thay vào phương trình ta được:

\(\frac{8}{3}\log_{\frac{x}{2}}2x+2\left(\frac{2}{\log_{\frac{x}{2}}2x}+\frac{1}{1+\log_2x}\right)=-\frac{14}{3}\)

Đặt \(\log_2x=a\left(a\in R\right)\)

Xét

\(\log_{\frac{x}{2}}2x=\log_{\frac{x}{2}}2+\log_{\frac{x}{2}}x=\frac{1}{\log_2\frac{x}{2}}+\frac{1}{\log_x\frac{x}{2}}=\frac{1}{\log_2x-1}+\frac{1}{1-\log_x2}=\frac{1}{a-1}+\frac{1}{1-\frac{1}{a}}=\frac{a+1}{a-1}\)

Thay vào pt ta được:

\(\frac{8}{3}\cdot\frac{a+1}{a-1}+2\left(2\cdot\frac{a-1}{a+1}+\frac{1}{a+1}\right)=-\frac{14}{3}\)

Giải ra ta được a=0 hoặc a=-23/17

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2^{-\frac{23}{17}}\end{matrix}\right.\)

a. \(2^{2\log_25+\log_{\frac{1}{2}}9}\) và \(\frac{\sqrt{626}}{6}\)

Ta có : \(2^{2\log_25+\log_{\frac{1}{2}}9}=2^{\log_225-\log_29}=2^{\log_2\frac{25}{9}}=\frac{25}{9}=\frac{\sqrt{625}}{9}< \frac{\sqrt{626}}{6}\)

           \(\Rightarrow2^{2\log_25+\log_{\frac{1}{2}}9}< \frac{\sqrt{626}}{6}\)

b. \(3^{\log_61,1}\) và \(7^{\log_60,99}\)

Ta có : \(\begin{cases}\log_61,1>0\Rightarrow3^{\log_61,1}>3^0=1\\\log_60,99< 0\Rightarrow7^{\log_60,99}< 7^0=1\end{cases}\)

             \(\Rightarrow3^{\log_61,1}>7^{\log_60,99}\)

c.  \(\log_{\frac{1}{3}}\frac{1}{80}\) và \(\log_{\frac{1}{2}}\frac{1}{15+\sqrt{2}}\)

Ta có : \(\begin{cases}\log_{\frac{1}{2}}\frac{1}{80}=\log_{3^{-1}}80^{-1}=\log_380< \log_381=4\\\log_{\frac{1}{2}}\frac{1}{15+\sqrt{2}}=\log_{2^{-1}}\left(15+\sqrt{2}\right)^{-1}=\log_2\left(15+\sqrt{2}\right)>\log_216=4\end{cases}\)

            \(\Rightarrow\log_{\frac{1}{3}}\frac{1}{80}< \log_{\frac{1}{2}}\frac{1}{15+\sqrt{2}}\)

d) Điều kiện x>0. Áp dụng công thức đổi cơ số, ta có :

\(\log_2x+\log_3x+\log_4x=\log_{20}x\)

\(\Leftrightarrow\log_2x+\frac{\log_2x}{\log_23}+\frac{\log_2x}{\log_24}=\frac{\log_2x}{\log_220}\)

\(\Leftrightarrow\log_2x\left(1+\frac{1}{\log_23}+\frac{1}{2}+\frac{1}{\log_220}\right)=0\)

\(\Leftrightarrow\log_2x\left(\frac{3}{2}+\log_22-\log_{20}2\right)=0\)

Ta có \(\frac{3}{2}+\log_22-\log_{20}2>\frac{3}{2}+0-1>0\)

Do đó, từ phương trình trên, ta phải có \(\log_2x=0\) hay \(x=2^0=1\)

Vậy nghiệm duy nhất của phương trình là \(x=1\)

c) Điều kiện x>0, đưa về cùng cơ số 5, ta có :

\(\log_5x^3+3\log_{25}x+\log_{\sqrt{25}}\sqrt{x^3}=\frac{11}{2}\)

\(\Leftrightarrow3\log_5x+3\log_{5^2}x+\log_{5^{\frac{3}{2}}}x^{\frac{3}{2}}=\frac{11}{2}\)

\(\Leftrightarrow3\log_5x+3\frac{1}{2}\log_5x+\frac{3}{2}.\frac{2}{3}\log_5x=\frac{11}{2}\)

\(\Leftrightarrow\frac{11}{2}\log_5x=\frac{11}{2}\)

\(\Leftrightarrow\log_5x=1\)

\(\Leftrightarrow x=5^1=5\) thỏa mãn

Vậy phương trình chỉ có 1 nghiệ duy nhất \(x=5\)

b) Điều kiện x>0. Đưa về cùng cơ số 2, ta có :

\(\log_2x+\log_{2^2}x+\log_{2^3}x=11\Leftrightarrow\log_2x+\frac{1}{2}\log_2x+\frac{1}{3}\log_2x=11\)

                                                 \(\Leftrightarrow\frac{11}{6}\log_2x=11\)

Do đó \(\log_2x=6\)

 và \(x=2^6=64\)

Vậy phương trình có nghiệm duy nhất là \(x=64\)