a: \(log\left(x-2\right)< 3\)

=>\(\left\{{}\begin{matrix}x-2>0\\log\left(x-2\right)< log9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-2>0\\x-2< 9\end{matrix}\right.\Leftrightarrow2< x< 11\)

b: \(log_2\left(2x-1\right)>3\)

=>\(\left\{{}\begin{matrix}2x-1>0\\log_2\left(2x-1\right)>log_29\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-1>0\\2x-1>9\end{matrix}\right.\Leftrightarrow2x-1>9\)

=>2x>10

=>x>5

c: \(log_3\left(-x-1\right)< =2\)

=>\(\left\{{}\begin{matrix}-x-1>0\\log_3\left(-x-1\right)< =log_39\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-x-1>0\\-x-1< =9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x>1\\-x< =10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< -1\\x>=-10\end{matrix}\right.\Leftrightarrow-10< =x< -1\)

d: \(log_2\left(2x-3\right)>=2\)

=>\(\left\{{}\begin{matrix}2x-3>0\\log_2\left(2x-3\right)>=log_24\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-3>0\\2x-3>=4\end{matrix}\right.\)

=>2x-3>=4

=>2x>=7

=>\(x>=\dfrac{7}{2}\)

e: \(log_3\left(2x-7\right)>2\)

=>\(\left\{{}\begin{matrix}2x-7>0\\log_3\left(2x-7\right)>log_39\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>\dfrac{7}{2}\\2x-7>9\end{matrix}\right.\)

=>2x-7>9

=>2x>16

=>x>8

a.

\(log\left(x-2\right)< 3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2>0\\x-2< 10^3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< 1002\end{matrix}\right.\) \(\Rightarrow2< x< 1002\)

b.

\(log_2\left(2x-1\right)>3\Leftrightarrow\left\{{}\begin{matrix}2x-1>0\\2x-1>2^3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{2}\\x>\dfrac{9}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{9}{2}\)

c.

\(log_3\left(-x-1\right)\le2\Rightarrow\left\{{}\begin{matrix}-x-1>0\\-x-1\le3^2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x< -1\\x\ge-10\end{matrix}\right.\) \(\Rightarrow-10\le x< -1\)

d.

\(log_2\left(2x-3\right)\ge2\Leftrightarrow\left\{{}\begin{matrix}2x-3>0\\2x-3\ge2^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x>\dfrac{7}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{7}{2}\)

e,

\(log_3\left(2x-7\right)>2\Leftrightarrow\left\{{}\begin{matrix}2x-7>0\\2x-7>3^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{7}{2}\\x>8\end{matrix}\right.\) \(\Rightarrow x>8\)

Lời giải:
a. ĐK: $x>2$
$\log(x-2)<3$

$\Leftrightarrow x-2< 10^3$

$\Leftrightarrow x< 1002$

Vậy $2< x< 1002$
b.  ĐK: $x> \frac{1}{2}$

$\log_2(2x-1)>3$

$\Leftrightarrow 2x-1> 2^3$

$\Leftrightarrow 2x> 9$

$\Leftrightarrow x> \frac{9}{2}$

Vậy $x> \frac{9}{2}$

c. ĐK: $x< -1$

$\log_3(-x-1)\leq 2$

$\Leftrightarrow -x-1\leq 3^2=9$

$\Leftrightarrow x+1\geq -9$

$\Leftrightarrow x\geq -10$

Vậy $-10\leq x< -1$

d. ĐK: $x> \frac{3}{2}$

$\log_2(2x-3)\geq 2$

$\Leftrightarrow 2x-3\geq 2^2=4$

$\Leftrightarrow x\geq \frac{7}{2}$

Vậy $x\geq \frac{7}{2}$

e. ĐK: $x> \frac{7}{2}$

$\log_3(2x-7)>2$
$\Leftrightarrow 2x-7> 3^2=9$
$\Leftrightarrow x> 8$

Vậy $x>8$

sao bạn k bấm máy tính. thi trắc nghiệm mà.

Đặt $\log_3x=a\Rightarrow a^2+\sqrt{a^2+1}-5=0\Leftrightarrow 4a^2+4\sqrt{a^2+1}-20=0$

$\Leftrightarrow (2\sqrt{a^2+1}+1)^2-25=0 \Rightarrow 2\sqrt{a^2+1}+1=5$

$\rightarrow a=\pm\sqrt{3}\Rightarrow x=3^a=3^{\sqrt{3}}$ hoặc $x=3^{-\sqrt{3}}$

Mình viết lại lời giải @@ Sao cứ bị lỗi công thức hoài nhỉ có cách nào sửa được không?

Đặt \(\log_3x=a\Rightarrow a^2+\sqrt{a^2+1}-5=0\Leftrightarrow4a^2+4\sqrt{a^2+1}-20=0\)

\(\Leftrightarrow\left(2\sqrt{a^2+1}+1\right)^2-25=0\Rightarrow2\sqrt{a^2+1}+1=5\) .Trường hợp $-5$ suy ra vô lý

\(\Rightarrow a=\pm\sqrt{3}\Leftrightarrow x=3^{\pm\sqrt{3}}\). Thử lại thấy đúng nên có hai nghiệm trên là nghiệm của PT

trong cac phan so sau :2/3 ;2/8 ;17/300 ;1/30.phan so thap phan la phan so 

ĐKXĐ: \(x>1\)

\(\Leftrightarrow\log_5\left(\dfrac{\ln x}{\ln3}\right)=\log_3\left(\dfrac{\ln x}{\ln5}\right)\)

\(\Leftrightarrow\log_5\left(\ln x\right)-\log_5\left(\ln3\right)=\log_3\left(\ln x\right)-\log_3\left(\ln5\right)\)

\(\Leftrightarrow\dfrac{\ln\left(\ln x\right)}{\ln5}-\log_5\left(\ln3\right)=\dfrac{\ln\left(\ln x\right)}{\ln3}-\log_3\left(\ln5\right)\)

\(\Leftrightarrow\ln\left(\ln x\right)\left(\dfrac{1}{\ln5}-\dfrac{1}{\ln3}\right)=\log_5\left(\ln3\right)-\log_3\left(\ln5\right)\)

\(\Leftrightarrow\ln\left(\ln x\right)=\dfrac{\log_5\left(\ln3\right)-\log_3\left(\ln5\right)}{\dfrac{1}{\ln5}-\dfrac{1}{\ln3}}=\dfrac{\ln3.\ln5\left[\log_5\left(\ln3\right)-\log_3\left(\ln5\right)\right]}{\ln3-\ln5}\)

\(\Rightarrow x=e^{e^{\frac{\ln{3}.\ln{5}[\log_{5}(\ln{3})-\log_{3}(\ln{5})]}{\ln{3}-\ln{5}}}}\)

a: \(log\left(x-5\right)< 2\)

=>\(\left\{{}\begin{matrix}x-5>0\\log\left(x-5\right)< log4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-5>0\\x-5< 4\end{matrix}\right.\Leftrightarrow5< x< 9\)

b: \(log_2\left(2x-3\right)>4\)

=>\(log_2\left(2x-3\right)>log_216\)

=>\(\left\{{}\begin{matrix}2x-3>0\\2x-3>16\end{matrix}\right.\)

=>2x-3>16

=>2x>19

=>\(x>\dfrac{19}{2}\)

c: \(log_3\left(2x+5\right)< =3\)

=>\(log_3\left(2x+5\right)< =log_327\)

=>\(\left\{{}\begin{matrix}2x+5>0\\2x+5< =27\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>-\dfrac{5}{2}\\x< =11\end{matrix}\right.\)

=>\(-\dfrac{5}{2}< x< =11\)

d: \(log_4\left(4x-5\right)>=2\)

=>\(log_4\left(4x-5\right)>=log_416\)

=>4x-5>=16 và 4x-5>0

=>4x>=21 và 4x>5

=>4x>=21

=>\(x>=\dfrac{21}{4}\)

e: \(log_3\left(1-3x\right)>3\)

=>\(log_3\left(1-3x\right)>log_327\)

=>\(\left\{{}\begin{matrix}1-3x>0\\1-3x>27\end{matrix}\right.\)

=>1-3x>27

=>\(-3x>26\)

=>\(x< -\dfrac{26}{3}\)

Điều kiện x>0. Nhận thấy x=2 là nghiệm. 

Nếu x>2 thì

\(\frac{x}{2}>\frac{x+2}{4}>1\); \(\frac{x+1}{3}>\frac{x+3}{5}>1\)

Suy ra 

\(\log_2\frac{x}{2}>\log_2\frac{x+2}{4}>\log_4\frac{x+2}{4}\)hay :\(\log_2x>\log_2\left(x+2\right)\)

\(\log_3\frac{x+1}{3}>\log_3\frac{x+3}{5}>\log_5\frac{x+3}{5}\) hay \(\log_3\left(x+1\right)>\log_5\left(x+3\right)\)

Suy ra vế trái < vế phải, phương trình vô nghiệm.

Đáp số x=2

a) Điều kiện: \(\left\{{}\begin{matrix}4x+2>0\\x-1>0\\x>0\end{matrix}\right.\)

Hay là: \(x>1\)

Khi đó biến đổi pương trình như sau:

\(\ln\dfrac{4x+2}{x-1}=\ln x\)

\(\Leftrightarrow\dfrac{4x+2}{x-1}=x\)

\(\Leftrightarrow4x+2=x\left(x-1\right)\)

\(\Leftrightarrow x^2-5x-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{5+\sqrt{33}}{2}\\x_2=\dfrac{5-\sqrt{33}}{2}\left(loại\right)\end{matrix}\right.\)

Vậy nghiệm của phương trình là: \(x=\dfrac{5+\sqrt{33}}{2}\)

b) Điều kiện: \(\left\{{}\begin{matrix}3x+1>0\\x>0\end{matrix}\right.\)

Hay là: \(x>0\)

Biến đổi phương trình như sau:

\(\log_2\left(3x+1\right)\log_3x-2\log_2\left(3x+1\right)=0\)

\(\Leftrightarrow\log_2\left(3x+1\right)\left(\log_3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\log_2\left(3x+1\right)=0\\\log_3x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=2^0\\x=3^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=9\end{matrix}\right.\)

Vậy nghiệm là x = 9.

c) Điều kiện: x > 0.

Khi đó biến đổi phương trình như sau:

\(2^{\log_3x^2}.5^{\log_3x}=400\)

\(\Leftrightarrow2^{2\log_3x}.5^{\log_3x}=400\)

\(\Leftrightarrow\left(2^2.5\right)^{\log_3x}=400\)

\(\Leftrightarrow20^{\log_3x}=20^2\)

\(\Leftrightarrow\log_3x=2\)

\(\Leftrightarrow x=3^2=9\) (thỏa mãn)

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: \(\left(\dfrac{1}{2};1\right)\)