255 Bài trắc nghiệm Hàm số mũ và Logarit cơ bản nâng cao cực hay có lời giải (P5)

Đáp án D

$$\begin{gathered} \mathrm{ĐK}: \mathrm{x}>0 \\ \sqrt{\mathrm{x}}.{\log }_3\sqrt{\mathrm{x}}=0 \\ \iff \sqrt{\mathrm{x}}=0 \mathrm{hoặc} {\log }_3\sqrt{\mathrm{x}}=0 \\ \iff \mathrm{x}=0 \left(\mathrm{L}\right) \mathrm{hoặc} \mathrm{x}=1 \left(\mathrm{TMĐK}\right) \end{gathered}$$

Đáp án B

$$\begin{gathered} ĐK: 2x-1>0\iff x>\frac{1}{2} \\ {\log }_2(2x-1)<3 \\ \iff 2x-1<8 \\ \iff x<\frac{9}{2} \\ Kết hợp điều kiện: \frac{1}{2}<x<\frac{9}{2} \end{gathered}$$

$$\begin{gathered} y={\log }_{x}5=\frac{\ln 5}{\ln x} \\ \Rightarrow y\text{'}=\ln 5.\frac{-{\displaystyle \frac{1}{x}}}{{\left(\ln x\right)}^2}=\frac{-\ln 5}{x{\left(\ln x\right)}^2}=\frac{-{\log }_{x}5}{x\ln x} \end{gathered}$$

Đáp án C

$$\begin{gathered} ĐK:\left\{\begin{array}{l}x\ne 1\\ x>0\end{array}\right. \\ {\log }_{x}3+{\log }_{3}x>2 \\ \iff \frac{1}{{\log }_{3}x}+{\log }_{3}x>2 (*) \\ ĐK: {\log }_{3}x>0\iff x>1 \\ (*)\iff {\log }_3^{2}x-2{\log }_{3}x+1>0 \\ \iff {({\log }_{3}x-1)}^2>0\iff {\log }_{3}x-1\ne 0 \\ \iff {\log }_{3}x\ne 1 \\ \iff x\ne 3 \\ Kết hợp ĐK: 1<x\ne 3 \end{gathered}$$

Đáp án B

$$\begin{gathered} Đặt 2^{x^2+1}=t (t>0) \\ \Rightarrow y=t+\frac{1}{t} \\ y\text{'}=1-\frac{1}{t^2}=\frac{t^2-1}{t^2} \\ y\text{'}=0 \iff t^2-1=0 \\ \iff t=1 \left(TM\right) hoặc t=-1 \left(L\right) \\ Ta thấy y\text{'} đổi dấu từ \text{'}\text{'}-\text{'}\text{'} \to \text{'}\text{'}+\text{'}\text{'} tại t=1 \\ {y}_{min} tại t=1 \\ \iff 2^{x^2+1}=1\iff x=0 \\ \Rightarrow y_{min }=y\left(0\right)=\frac{5}{2} \end{gathered}$$

Đáp án C

$$\begin{gathered} \mathrm{Nghiệm} \mathrm{của} \mathrm{hệ} \mathrm{thỏa} \mathrm{mãn} \mathrm{phơng} \mathrm{trình}: \\ {\mathrm{T}}^2-\frac{10}{3}\mathrm{T}+1=0 \\ \iff \mathrm{T}=3 \mathrm{hoặc} \mathrm{T}=\frac{1}{3} \\ \mathrm{TH}1: \\ \left\{\begin{array}{l}{2}^{\mathrm{x}}=3\\ 3^{\mathrm{y}}=\frac{1}{3}\end{array}\iff \left\{\begin{array}{l}\mathrm{x}={\log }_{2}3\\ \mathrm{y}=-1\end{array}\right.\right. \\ \mathrm{TH}2: \\ \left\{\begin{array}{l}{2}^{\mathrm{x}}=\frac{1}{3}\\ 3^{\mathrm{y}}=3\end{array}\iff \left\{\begin{array}{l}\mathrm{x}={\log }_2\left(\frac{1}{3}\right)\\ \mathrm{y}=1\end{array}\right.\right. \end{gathered}$$

Đáp án D

${\log }_3{\mathrm{a}}^2{\mathrm{b}}^3={\log }_3{\mathrm{a}}^2+{\log }_3{\mathrm{b}}^3=2{\log }_3\mathrm{a}+3{\log }_3\mathrm{b}$

Đáp án B

$$\begin{gathered} \mathrm{Ta} \mathrm{có}: \mathrm{y}\text{'}=3.2^{3\mathrm{x}-1}.\ln 2>0 \mathrm{với} \forall \mathrm{x}\in \mathrm{ℝ} \\ \Rightarrow \mathrm{y}\uparrow /\mathrm{ℝ} \end{gathered}$$

Đáp án C

$$\begin{gathered} ĐK: 3x-1>0\iff x>\frac{1}{3} \\ {\log }_2\left(3x-1\right)=\sqrt{2} \\ \iff 3x-1=2^{\sqrt{2}} \\ \iff x=\frac{1+2^{\sqrt{2}}}{3} \left(TMĐK\right) \end{gathered}$$

Đáp án A

$$\begin{gathered} {\left(\frac{1}{2}\right)}^{x^2}>2^x \\ \iff -x^2>x \\ \iff x^2+x<0 \\ \iff -1<x<0 \end{gathered}$$

Đáp án C

$$\begin{gathered} ĐK: \left\{\begin{array}{l}2x-1>0\\ x>0\end{array}\right.\iff \left\{\begin{array}{l}x>\frac{1}{2}\\ x>0\end{array}\right.\iff x>\frac{1}{2} \\ {\log }_{2-\sqrt{3}}(2x-1)>{\log }_{2-\sqrt{3}}x \\ \iff 2x-1<x \\ \iff x<1 \\ Kết hợp ĐK: \frac{1}{2}<x<1 \end{gathered}$$

Đáp án B

$$\begin{gathered} ĐK: 4^{x-x^2}-1\ne 0 \\ \iff 4^{x-x^2}\ne 1 \\ \iff x-x^2\ne 0 \\ \iff \left\{\begin{array}{l}x\ne 0\\ x\ne 1\end{array}\right. \end{gathered}$$

Đáp án D

$$\begin{gathered} Xét đáp án A: \\ {\left(\frac{1}{3}\right)}^{\frac{1}{x}}=3^{\frac{-1}{x}}=3^{-\left(x^{-1}\right)}\ne 3^x \\ Xét đáp án B: \\ {\left(\frac{1}{3}\right)}^x=\frac{1^x}{3^x} \\ Xét đáp án C: \\ {\left(\frac{1}{9}\right)}^{\frac{2}{x}}={\left(\frac{1}{3}\right)}^{2.\frac{2}{x}}={\left(\frac{1}{3}\right)}^{\frac{4}{x}} \\ Xét đáp án D: \\ {\left(\frac{1}{27}\right)}^{\frac{1}{3x}}={\left(\frac{1}{3}\right)}^{3.\frac{1}{3x}}={\left(\frac{1}{3}\right)}^{\frac{1}{x}}( đúng ) \end{gathered}$$

Đáp án D

$$\begin{gathered} A,B đúng khi 0<x\ne 1 \\ C đúng khi x>0 \end{gathered}$$

Đáp án D

$$\begin{gathered} TH1: a>1 \\ {\log }_{a}b>1\iff \left\{\begin{array}{l}b>0\\ b>a\end{array}\right. \\ \Rightarrow b>a>1 \\ TH2: 0<a<1 \\ {\log }_{a}b>1\iff \left\{\begin{array}{l}b>0\\ b<a\end{array}\right. \\ \Rightarrow 0<b<a<1 \end{gathered}$$

Đáp án B

$$\begin{gathered} ĐK: {\log }_5\left(\frac{2x-3}{1-x}\right)\ge 0 \\ \iff \frac{2x-3}{1-x}\ge 1 \\ \iff \frac{2x-3}{1-x}-1\ge 0 \\ \iff \frac{3x-4}{1-x}\ge 0 \\ \iff 1<x\le \frac{4}{3} \end{gathered}$$

Đáp án D

$$\begin{gathered} 3^{{\mathrm{x}}^2-2\mathrm{x}}=4^{\mathrm{x}-2} \\ \iff {\log }_3\left(3^{{\mathrm{x}}^2-2\mathrm{x}}\right)={\log }_3\left(4^{\mathrm{x}-2}\right) \\ \iff {\mathrm{x}}^2-2\mathrm{x}=(\mathrm{x}-2){\log }_{3}4 \\ \iff (\mathrm{x}-2)(\mathrm{x}-{\log }_{3}4)=0 \\ \iff \mathrm{x}-2=0 \mathrm{hoặc} \mathrm{x}-{\log }_{3}4=0 \\ \iff \mathrm{x}=2 \mathrm{hoặc} \mathrm{x}={\log }_{3}4 \end{gathered}$$

Ta thấy A chỉ đúng khi x > 1

$$\begin{gathered} \mathrm{B}: \iff 5^x+3^x-9^x<0 \\ \iff {\left(\frac{5}{9}\right)}^x+{\left(\frac{3}{9}\right)}^x-1<0 \\ B đúng khi x>1 \end{gathered}$$

$$\begin{gathered} \mathrm{C}: {\left(\frac{3}{2}\right)}^{\mathrm{x}}+2^{\mathrm{x}}>0 \\ C đúng khi x>1 \end{gathered}$$

$$\begin{gathered} D: ta có {(x-1)}^2\ge 0 với\forall x\in \mathrm{ℝ} \\ \iff x^2-2x+1\ge 0 \\ \iff 2x-1\le x^2 \\ \iff 2^{2x-1}\le 2^{x^2} \\ dấu \text{'}\text{'}=\text{'}\text{'} xảy ra khi x=1 \end{gathered}$$

Đáp án A

$$\begin{gathered} ĐK: \left\{\begin{array}{l}0<1-2x\ne 1\\ 1-3x>0\end{array}\right.\iff \left\{\begin{array}{l}0\ne x<\frac{1}{2}\\ x<\frac{1}{3}\end{array}\iff \left\{\begin{array}{l}x\ne 0\\ x<\frac{1}{3}\end{array}\right.\right. \\ {\log }_{1-2x}(1-3x)=-1 \\ \iff 1-3x={(1-2x)}^{-1}\iff (1-3x)(1-2x)=1 \\ \iff 6x^2-5x=0 \\ \iff x=0\left(L\right) hoặc x=\frac{5}{6}\left(L\right) \end{gathered}$$

Đáp án C

${108}^{\mathrm{x}}={\left(2^2.3^3\right)}^{\mathrm{x}}={\left(2^{\mathrm{x}}\right)}^2.{\left(3^{\mathrm{x}}\right)}^3={\mathrm{a}}^2.{\mathrm{b}}^3$

$$\begin{gathered} \mathrm{F}\text{'}=1-\frac{81}{{\mathrm{t}}^2}=\frac{{\mathrm{t}}^2-81}{{\mathrm{t}}^2} \\ \mathrm{F}\text{'}=0\iff {\mathrm{t}}^2-81=0\iff \mathrm{t}=9 ( \mathrm{t}=-9 \mathrm{bị} \mathrm{loại} ) \\ \mathrm{Ta} \mathrm{có} \mathrm{F}\left(1\right)=82; \mathrm{F}\left(9\right)=18; \mathrm{F}\left(81\right)=82 \\ \Rightarrow \min \mathrm{F} =18 \mathrm{và} \max \mathrm{F}=82 \end{gathered}$$

Đáp án D

$$\begin{gathered} t={\left(2-\sqrt{3}\right)}^x\iff {\left(2+\sqrt{3}\right)}^x=\frac{1}{t} \\ \Rightarrow {(7+4\sqrt{3})}^x={\left(2+\sqrt{3}\right)}^{2x}=\frac{1}{t^2} \end{gathered}$$

Đáp án B

$$\begin{gathered} ĐK: \left\{\begin{array}{l}3x^2-6x+2>0\\ 1-2x>0\end{array}\iff \left\{\begin{array}{l}x>\frac{3+\sqrt{3}}{3} hoặc x<\frac{3-\sqrt{3}}{3}\\ x<\frac{1}{2}\end{array}\right.\right. \\ \iff x<\frac{3-\sqrt{3}}{2} \\ {\log }_6(3x^2-6x+2)={\log }_6(1-2x) \\ \iff 3x^2-6x+2=1-2x \\ \iff 3x^2-4x+1=0 \\ \iff x=1 \left(L\right) hoặc x=\frac{1}{3}\left(TMĐK\right) \end{gathered}$$

Đáp án C

$$\begin{gathered} \left(3^x-m\right)\left(3^x+2\right)=0 \\ PT có nghiệm khi PT 3^x-m=0 do PT 3^x+2=0 vô nghiệm \\ \iff m>0 \end{gathered}$$

Đáp án C

$$\begin{gathered} ĐK: {\log }_{4}x\ne 0\iff 0<x\ne 1 \\ PT\iff 2^x.3^{2x}-5^{x^2-x}=0 \\ \iff {18}^x-5^{x^2-x}=0 \\ \iff 5^{x^2-x}={18}^x \\ \iff {\log }_5\left(5^{x^2-x}\right)={\log }_5\left({18}^x\right) \\ \iff x^2-x=x{\log }_{5}18 \\ \iff x\left[\right(x-1)-{\log }_{5}18]=0 \\ \iff x=0 \left(L\right) hoặc x=1+{\log }_{5}18 \left(TMĐK\right) \end{gathered}$$

Đáp án D

$$\begin{gathered} {\log }_2\left(a{b}^2{c}^3\right)={\log }_{2}a+{\log }_2{b}^2+{\log }_2{c}^3 \\ ={\log }_{2}a+2{\log }_{2}b+3{\log }_{2}c \end{gathered}$$

Đáp án A

$$\begin{gathered} BPT\iff 3x^2-1\le 0 do 0< 5-2\sqrt{6}<1 \\ \iff \frac{-1}{\sqrt{3}}\le x\le \frac{1}{\sqrt{3}} \end{gathered}$$

Đáp án B

$$\begin{gathered} y\text{'}=\frac{2x+1}{(x^2+x+1)\ln 3}-\frac{1}{x\ln 3}=\frac{x(2x+1)-(x^2+x+1)}{x(x^2+x+1).\ln 3} \\ =\frac{x^2-1}{x(x^2+x+1).\ln 3} \\ y\text{'}=0\iff x=1 hoặc x=-1 \left(L\right) \\ Lập BBT ta thấy y_{min}=y\left(1\right)=1 khi x>0 \end{gathered}$$

Đáp án D

$$\begin{gathered} TXĐ: D=\mathrm{ℝ} \\ f\text{'}\left(x\right)=2x.{\left(2+\sqrt{5}\right)}^{x^2-1}\ln (2+\sqrt{5}) \\ f\text{'}\left(x\right)=0\iff x=0 \\ ta thấy f\text{'}\left(x\right)>0\iff x>0 và f\text{'}\left(x\right)<0\iff x<0 \\ \Rightarrow f\left(x\right) \uparrow /(0;+\infty ) và f\left(x\right)\downarrow /(-\infty ;0) \end{gathered}$$

Đáp án B

$$\begin{gathered} PT\iff 2x-1=\frac{1}{2}x \\ \iff 4x-2=x\iff x=\frac{2}{3} \end{gathered}$$

Đáp án C

$$\begin{gathered} ĐK: \left\{\begin{array}{l}2x^2-1>0\\ x>0\end{array}\right.\iff \left\{\begin{array}{l}x<\frac{-1}{\sqrt{2}} hoặc x>\frac{1}{\sqrt{2}}\\ x>0\end{array}\right.\iff x>\frac{1}{\sqrt{2}} \\ BPT\iff -{\log }_4(2x^2-1)>-{\log }_{4}x \\ \iff {\log }_4(2x^2-1)<{\log }_{4}x \\ \iff 2x^2-1<x \\ \iff 2x^2-x-1<0 \\ \iff \frac{-1}{2}<x<1 \\ Kết hợp điều kiện: \frac{1}{\sqrt{2}}<x<1 \end{gathered}$$