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

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