Giải SBT Toán học 11 CTST Bài 2: Phép tính lôgarit có đáp án

a) ${\log }_{9}\frac{1}{81}=lo{g}_{9}{(9)}^{-2}=-2$;

b) log 10 000 = log 10$^{4}$= 4;

c) log 0,001 = log(10)$^{–}$$^{3}$= – 3;

d) log$_{0{,}7}$1 = 0;

e) $lo{g}_{5}\sqrt[4]{5}={\log }_{{}_{5}}{(5)}^{\frac{1}{4}}=\frac{1}{4}$ ;

g) log$_{0{,}5}$0,125 = log$_{0{,}5}$0,5$^{3}$ = 3.

a) ${3}^{{\log }_{3}5}=5$ ;

b) e$^{ln 3}$= 3;

c) ${7}^{2{\log }_{7}8}={[{7}^{lo{g}_{7}8}]}^{2}={8}^{2}=64$ ;

d) ${2}^{lo{g}_{2}3+{\log }_{2}5}={2}^{lo{g}_{2}3}{.2}^{{\log }_{2}5}=3.5=15$ ;

e) ${4}^{lo{g}_{2}(\frac{1}{5})}={({2}^{2})}^{lo{g}_{2}(\frac{1}{5})}$

$={[{2}^{lo{g}_{2}\frac{1}{5}}]}^{2}={[\frac{1}{5}]}^{2}=\frac{1}{25}$

g) 0,001$^{log 2}$ = ${({10}^{-3})}^{log2}$ = ${({10}^{log2})}^{-3}$ = ${2}^{-3}$ = $\frac{1}{8}$ .

a) ${\log }_{3}\frac{9}{10}+{\log }_{3}30={\log }_{3}(\frac{9}{10}.30)={\log }_{3}{3}^{3}=3$ ;

b) ${\log }_{5}75-{\log }_{5}3$$={\log }_{5}\frac{75}{3}={\log }_{5}25$$={\log }_{5}{5}^{2}=2$ ;

c) ${\log }_{3}\frac{5}{9}-2{\log }_{3}\sqrt{5}$$={\log }_{3}\frac{5}{9}-{\log }_{3}{(\sqrt{5})}^{2}$

$={\log }_{3}\frac{5}{9}-{\log }_{3}5$$={\log }_{3}(\frac{5}{9}:5)$$={\log }_{3}{3}^{-2}=-2$

d) $4{\log }_{12}2+2{\log }_{12}3={\log }_{12}{2}^{4}+{\log }_{12}{3}^{2}$

$={\log }_{12}({2}^{4}{.3}^{3})={\log }_{12}{(4.3)}^{2}$$={\log }_{12}{12}^{2}=2$

e) $2{\log }_{5}2-{\log }_{5}4\sqrt{10}+{\log }_{5}\sqrt{2}$

$={\log }_{5}{2}^{2}-{\log }_{5}4\sqrt{10}+{\log }_{5}\sqrt{2}$= ${\log }_{5}4-{\log }_{5}4\sqrt{10}+{\log }_{5}\sqrt{2}$

$={\log }_{5}\frac{4\sqrt{2}}{4\sqrt{10}}={\log }_{5}\frac{1}{\sqrt{5}}$$={\log }_{5}{5}^{-\frac{1}{2}}=-\frac{1}{2}$;

g) ${\log }_{3}\sqrt{3}-{\log }_{3}\sqrt[3]{9}+2{\log }_{3}\sqrt[4]{27}$

$={\log }_{3}{3}^{\frac{1}{2}}-{\log }_{3}{3}^{\frac{2}{3}}+2{\log }_{3}{3}^{\frac{3}{4}}$
$=\frac{1}{2}-\frac{2}{3}+2.\frac{3}{4}=\frac{4}{3}$

a) ${\log }_{8}\frac{1}{32}=\frac{{\log }_{2}\frac{1}{32}}{{\log }_{2}8}=\frac{{\log }_{2}{2}^{-5}}{{\log }_{2}{2}^{3}}=-\frac{5}{3}$ ;

b) ${\log }_{5}3.{\log }_{3}5={\log }_{5}3\frac{1}{{\log }_{5}3}=1$ ;

c) ${2}^{\frac{1}{{\log }_{5}2}}={2}^{lo{g}_{2}5}=5$ ;

d) ${\log }_{27}25.{\log }_{5}81=\frac{{\log }_{3}25}{{\log }_{3}27}\cdot \frac{{\log }_{3}81}{{\log }_{3}5}$

$=\frac{{\log }_{3}{5}^{2}}{{\log }_{3}{3}^{3}}\cdot \frac{{\log }_{3}{3}^{4}}{{\log }_{3}5}=\frac{2{\log }_{3}5}{3}\cdot \frac{4}{{\log }_{3}5}$$=\frac{8}{3}$

a) ${\log }_{3}5.{\log }_{5}7.{\log }_{7}9$

$={\log }_{3}5.\frac{{\log }_{3}7}{{\log }_{3}5}.\frac{{\log }_{3}9}{{\log }_{3}7}$

$=lo{g}_{3}{3}^{2}=2$

b) ${\log }_{2}\frac{1}{25}.{\log }_{3}\frac{1}{32}.{\log }_{5}\frac{1}{27}={\log }_{2}{5}^{-2}.{\log }_{3}{2}^{-5}.{\log }_{5}{3}^{-3}$

<=> (–2) log$_{2}$5. (–5)log$_{3}$ 2. (–3) . (–3) log$_{5}$3

<=> –30 log$_{2}$ 5 . log$_{2}$3 . log$_{5}$3 log$_{7}$21

$\Leftrightarrow -30{\log }_{2}5\ .\ \frac{{\log }_{2}2}{{\log }_{2}3}\ .\ \frac{{\log }_{2}3}{{\log }_{2}5}=-30$

a) log$_{7}$21 = 1,5646;

Ta nhập máy tính:

Ta được kết quả:

Làm tròn đến chữ số thập phân thứ tư 1,5646.

b) log 2,25 = 0,3522;

Ta nhập máy tính:

Ta được kết quả:

Làm tròn đến chữ số thập phân thứ tư 0,3522.

c) = 1,3195;

Ta nhập máy tính:

Ta được kết quả:

Làm tròn đến chữ số thập phân thứ tư 1,3195.

d) log$_{0{,}5}$ 3 + log$_{5}$0,3 = –2,333.

Ta nhập máy tính:

Ta được kết quả:

Làm tròn đến chữ số thập phân thứ tư – 2,333.

a) log$_{2}$45 = log$_{2}$ 3$^{2}$.5

= 2log$_{2}$3 + log$_{2}$5 = 2a + b;

b) ${\log }_{2}\frac{\sqrt{15}}{6}={\log }_{2}\sqrt{15}-{\log }_{2}6$$=\frac{1}{2}{\log }_{2}15-{\log }_{2}(2\ .\ 3)$

$=\frac{1}{2}{\log }_{2}(3\ .\ 5)\ -(\ {\log }_{2}2\ +\ {\log }_{2}3)$$=\frac{1}{2}({\log }_{2}3+{\log }_{2}5)-(1+{\log }_{2}3)$

$=\frac{1}{2}(a+b)-(1+a)$$=-\frac{a}{2}+\frac{b}{2}-1$

c) ${\log }_{3}20=\frac{{\log }_{3}20}{{\log }_{2}3}$$=\frac{{\log }_{2}({2}^{2}\ .\ 5)}{{\log }_{3}3}$

$=\frac{2{\log }_{2}2+{\log }_{2}5}{{\log }_{2}3}$$=\frac{2+b}{a}$

a) log(xyz) = log x + log y + log z = a + b + c;

b) $\log \frac{{x}^{3}\sqrt[3]{y}}{100\sqrt{z}}=\log ({x}^{3}\sqrt[3]{y})-\log (100\sqrt{z})$

$=\log ({x}^{3}{y}^{\frac{1}{3}})-\log ({10}^{2}{z}^{\frac{1}{2}})$$=3\log x+\frac{1}{3}\log y-2-\frac{1}{2}\log z$

$=3a+\frac{1}{3}b-\frac{1}{2}c-2$

c) ${\log }_{z}(x{y}^{2})=\frac{\log (x{y}^{2})}{\log z}$

$=\frac{\log xy+2\log }{\log z}=\frac{a+2b}{c}$

Ta có:

$a={\log }_{2}3=\frac{1}{{\log }_{3}2}$$\Rightarrow {\log }_{3}2=\frac{1}{a}$

$b={\log }_{3}15={\log }_{3}(3\ .\ 5)$

$={\log }_{3}3+{\log }_{3}5$$=1+{\log }_{3}5$

$\Rightarrow {\log }_{3}5=b-1$

${\log }_{30}18=\frac{{\log }_{3}18}{{\log }_{3}30}$$=\frac{{\log }_{3}(2\ .\ {3}^{2})}{{\log }_{3}(2\ .\ 3\ .5)}$

$=\frac{{\log }_{3}2+{\log }_{3}{3}^{2}}{{\log }_{3}2+{\log }_{3}3+{\log }_{3}5}$$=\frac{{\log }_{3}2+2}{{\log }_{3}2+1+{\log }_{3}5}$

$=\frac{\frac{1}{a}+2}{\frac{1}{a}+1+b-1}$$=\frac{2a+1}{ab+1}$

Vậy ${\log }_{30}18$$=\frac{2a+1}{ab+1}$