Câu 76:

Thay t=3754 và N(t)=65 vào N(t), ta được:

\(100\left(\dfrac{1}{2}\right)^{\dfrac{3754}{A}}=65\)

=>\(\left(\dfrac{1}{2}\right)^{\dfrac{3754}{A}}=\dfrac{65}{100}=\dfrac{13}{20}\)

=>\(\dfrac{3754}{A}=log_{\dfrac{1}{2}}\left(\dfrac{13}{20}\right)\)

=>\(A\simeq6040,34\)

Đặt N(t)=79

=>\(100\left(\dfrac{1}{2}\right)^{\dfrac{t}{6040,34}}=79\)

=>\(\left(\dfrac{1}{2}\right)^{\dfrac{t}{6040,34}}=\dfrac{79}{100}\)

=>\(\dfrac{t}{6040,34}=log_{\dfrac{1}{2}}\left(0,79\right)\)

=>\(t\simeq2054\)

=>Chọn D

Câu 73:

Đặt Q(t)=90%=0,9

=>\(1\left(1-e^{-t\sqrt{2}}\right)=0,9\)

=>\(1-e^{-t\sqrt{2}}=0,9\)

=>\(e^{-t\sqrt{2}}=0,1\)

=>\(-t\sqrt{2}=log_e0,1\)

=>\(t\simeq1,63\)

=>Chọn C

Câu 71:

\(a=log_915=log_{3^2}15=\dfrac{1}{2}\cdot log_315\)

\(b=log_{27}45=log_{3^3}45=\dfrac{1}{3}\cdot log_345=\dfrac{1}{3}\cdot log_3\left(3\cdot15\right)\)

\(=\dfrac{1}{3}\cdot\left(log_33+log_315\right)\)

\(=\dfrac{1}{3}\left(1+log_315\right)=\dfrac{1}{3}+\dfrac{1}{3}\cdot log_315\)

\(2a+3b=log_315+3\cdot\dfrac{1}{3}\cdot\left(1+log_315\right)\)

\(=2\cdot log_315+1\)

=>Loại B và D

\(3b-2a=3\left(\dfrac{1}{3}+\dfrac{1}{3}\cdot log_315\right)-2\cdot\dfrac{1}{2}\cdot log_315\)

\(=1+log_315-log_315=1\)

=>Chọn A

Câu 69:

\(log_{\sqrt[3]{a}}\left(\dfrac{b^2}{\sqrt[3]{c^2}}\right)\)

\(=log_{a^{\dfrac{1}{3}}}\left(\dfrac{b^2}{c^{\dfrac{2}{3}}}\right)\)

\(=3\cdot log_a\left(\dfrac{b^2}{c^{\dfrac{2}{3}}}\right)\)

\(=3\cdot log_ab^2-3\cdot log_ac^{\dfrac{2}{3}}\)

\(=6\cdot log_ab-2\cdot log_ac=6m-2n\)

=>Chọn B

Câu 61:

\(log_2\left(\dfrac{2a^3}{b}\right)\)

\(=log_22a^3-log_2b\)

\(=log_22+log_2a^3-log_2b\)

\(=1+3\cdot log_2a-log_2b\)

=>Chọn C

Câu 62:

\(log_{27}\left(\dfrac{\sqrt{x}}{y}\right)^3\)

\(=log_{27}\left(\dfrac{x^{\dfrac{1}{2}}}{y}\right)^3\)

\(=3\cdot log_{3^3}\left(\dfrac{x^{\dfrac{1}{2}}}{y}\right)\)

\(=3\cdot\dfrac{1}{3}\cdot log_3\left(\dfrac{x^{\dfrac{1}{2}}}{y}\right)\)

\(=log_3\left(\dfrac{x^{\dfrac{1}{2}}}{y}\right)=log_3x^{\dfrac{1}{2}}-log_3y=\dfrac{1}{2}a-b\)

=>Chọn C

Câu 63

\(log_{a^2}\left(ab\right)=\dfrac{1}{2}\cdot log_aab\)

\(=\dfrac{1}{2}\left(log_aa+log_ab\right)\)

\(=\dfrac{1}{2}\left(1+log_ab\right)=\dfrac{1}{2}+\dfrac{1}{2}\cdot log_ab\)

=>Chọn B

Câu 66:

\(log_256=log_2\left(2^3\cdot7\right)=3+log_27=3+a\)

=>Chọn D

Chọn B

Kẻ DE vuông góc AB và DF vuông góc AC (D là đỉnh cột điện)

\(EA=AD.cos\alpha=3,75.\dfrac{3}{5}=2,25\)

\(\Rightarrow DF=BE=1,75+2,25=4\)

\(\Rightarrow DC=\dfrac{DF}{sin\alpha}=\dfrac{4}{\sqrt{1-\left(\dfrac{3}{5}\right)^2}}=5\)

ĐKXĐ: \(x\ne\dfrac{\pi}{4}+k'\dfrac{\pi}{2}\)

pt\(\Leftrightarrow2x=\dfrac{\pi}{4}-x+k\pi\)

\(\Rightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\)

\(0\le\dfrac{\pi}{12}+\dfrac{k\pi}{3}\le3\pi\Rightarrow0\le k\le8\)

Loại nghiệm: \(\dfrac{\pi}{4}+\dfrac{k'\pi}{2}\ne\dfrac{\pi}{12}+\dfrac{k\pi}{3}\Rightarrow\dfrac{1+3k'}{2}\ne k\)

Để k nguyên \(\Rightarrow k'\) lẻ \(\Rightarrow k'=2n+1\Rightarrow k\ne\dfrac{3\left(2n+1\right)+1}{2}=3n+2\)

Tới đây ít số thì loại trực triếp \(k=\left\{2;5;8\right\}\) còn nhiều số thì biện luận do \(k\ne3n+2\) nên k có dạng \(k=3n\) hoặc \(k=3n+1\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+\dfrac{3n.\pi}{3}=\dfrac{\pi}{12}+n\pi\\x=\dfrac{\pi}{12}+\dfrac{\left(3n+1\right)\pi}{3}=\dfrac{5\pi}{12}+n\pi\end{matrix}\right.\)

ĐKXĐ: \(\left\{{}\begin{matrix}2x\ne\dfrac{\Omega}{2}+k\Omega\\\dfrac{\Omega}{4}-x\ne\dfrac{\Omega}{2}+k\Omega\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\ne\dfrac{\Omega}{4}+\dfrac{k\Omega}{2}\\x\ne-\dfrac{\Omega}{4}-k\Omega\end{matrix}\right.\)

=>\(x\ne\dfrac{\Omega}{4}+\dfrac{k\Omega}{2}\)

\(tan2x=tan\left(\dfrac{\Omega}{4}-x\right)\)

=>\(2x=\dfrac{\Omega}{4}-x+k\Omega\)

=>\(3x=\dfrac{\Omega}{4}+k\Omega\)

=>\(x=\dfrac{\Omega}{12}+\dfrac{k\Omega}{3}\)

mà \(x\in\left[0;3\Omega\right]\)

nên \(\dfrac{\Omega}{12}+\dfrac{k\Omega}{3}\in\left[0;3\Omega\right]\)

=>\(\dfrac{1}{12}+\dfrac{k}{3}\in\left[0;3\right]\)

=>\(\dfrac{k}{3}\in\left[-\dfrac{1}{12};\dfrac{35}{12}\right]\)

=>\(k\in\left[-\dfrac{1}{4};\dfrac{35}{4}\right]\)

mà k nguyên

nên \(k\in\left\{0;1;2;3;4;5;6;7;8\right\}\)

=>\(\dfrac{\Omega}{12}+\dfrac{k\Omega}{3}\in\left\{\dfrac{1}{12}\Omega;\dfrac{5}{12}\Omega;\dfrac{3}{4}\Omega;\dfrac{13}{12}\Omega;\dfrac{17}{12}\Omega;\dfrac{7}{4}\Omega;\dfrac{25}{12}\Omega;\dfrac{29}{12}\Omega;\dfrac{11}{4}\Omega\right\}\)

=>\(x\in\left\{\dfrac{1}{12}\Omega;\dfrac{5}{12}\Omega;\dfrac{3}{4}\Omega;\dfrac{13}{12}\Omega;\dfrac{17}{12}\Omega;\dfrac{7}{4}\Omega;\dfrac{25}{12}\Omega;\dfrac{29}{12}\Omega;\dfrac{11}{4}\Omega\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{\dfrac{1}{12}\Omega;\dfrac{5}{12}\Omega;\dfrac{13}{12}\Omega;\dfrac{17}{12}\Omega;\dfrac{25}{12}\Omega;\dfrac{29}{12}\Omega\right\}\)

Tổng các nghiệm là:

\(\Omega\left(\dfrac{1}{12}+\dfrac{5}{12}+\dfrac{13}{12}+\dfrac{17}{12}+\dfrac{25}{12}+\dfrac{29}{12}\right)=\dfrac{15}{2}\Omega\)