Đáp án đúng là: B

Với a > 0, a ≠ 2 ta có:  ${\log }_{\frac{a}{2}}\left(\frac{a^2}{4}\right)={\log }_{\frac{a}{2}}\left(\frac{a^2}{2^2}\right)={\log }_{\frac{a}{2}}{\left(\frac{a}{2}\right)}^2=2.$

Đáp án đúng là: C
Với a > 0, a ≠ 1 ta có:
${\log }_a\sqrt{a\sqrt{a}}={\log }_a{\left(a.a^{\frac{1}{2}}\right)}^{\frac{1}{2}}={\log }_a{\left(a^{\frac{3}{2}}\right)}^{\frac{1}{2}}$$={\log }_a{a}^{\frac{3}{2}.\frac{1}{2}}=lo{g}_a{a}^{\frac{3}{4}}=\frac{3}{4}.$

Đáp án đúng là: A

Với a > 0 ta có: ${\log }_2\left(\frac{8}{a}\right)={\log }_{2}8-{\log }_{2}a={\log }_2{2}^3-{\log }_{2}a$

            = 3log₂2 – log₂ a = 3 – log₂ a.

Đáp án đúng là: B

Với a > 0, b > 0, c > 0, a ≠ 1 ta có:

log_a(b²c³) = log_ab² + log_ac³ = 2log_ab + 3log_ac = 2.2 + 3.3 = 13.

Đáp án đúng là: D

Với a > 0 ta có:

   ln(9a) – ln(3a) = ln(3.3a) – ln(3a)

= ln3 + ln(3a) – ln(3a) = ln3.

Đáp án đúng là: A

Với a > 0, b > 0 ta có: ${\log }_2\left(\frac{2a^3}{b}\right)={\log }_2\left(2a^3\right)-{\log }_{2}b$

= log₂2 + log₂a³ – log₂b = 1 + 3log₂a – log₂b.

Đáp án đúng là: D

Với a > 0, a ≠ 1 và b > 0 ta có: ${\log }_{a^2}\left(ab\right)=\frac{1}{2}{\log }_a\left(ab\right)=\frac{1}{2}\left({\log }_{a}a+{\log }_{a}b\right)$

$=\frac{1}{2}\left(1+lo{g}_{a}b\right)=\frac{1}{2}+\frac{1}{2}{\log }_{a}b.$

Đáp án đúng là: D

Nếu log₂3 = a thì  ${\log }_{6}9=\frac{{\log }_{2}9}{{\log }_{2}6}=\frac{{\log }_2{3}^2}{{\log }_{2}3+lo{g}_{2}2}$ $=\frac{2{\log }_{2}3}{{\log }_{2}3+1}=\frac{2a}{a+1}.$

 

Đáp án đúng là: A
Với a > 0, b > 0, a ≠ 1 và logab = 5 thì

${\log }_{a^{2}b}\left(a{b}^2\right)=\frac{{\log }_a\left(a{b}^2\right)}{{\log }_a\left(a^{2}b\right)}=\frac{{\log }_{a}a+{\log }_a{b}^2}{{\log }_a{a}^2+{\log }_{a}b}$ $=\frac{1+2{\log }_{a}b}{2+lo{g}_{a}b}=\frac{1+2\cdot 5}{2+5}=\frac{11}{7}.$

Đáp án đúng là: D

Với a > 0, b > 0 ta có:

a² + b² = 7ab hay a² + 2ab + b² = 9ab ⇒ (a + b)² = 9ab.

 $\Rightarrow a+b=\sqrt{9ab}\Rightarrow a+b=3{\left(ab\right)}^{\frac{1}{2}}$ (Vì a > 0, b > 0).

Xét:  $\log \left(a+b\right)=\log \left[3{\left(ab\right)}^{\frac{1}{2}}\right]$

                     $=\log 3+\log {\left(ab\right)}^{\frac{1}{2}}$

                     $=\log 3+\frac{1}{2}\left(\log a+\log b\right).$

a) ${\log }_{\sqrt{2}}8={\log }_{2^{\frac{1}{2}}}{2}^3=2{\log }_2{2}^3$$=\mathrm{2.3.}{\log }_{2}2=6.1=6.$

b) ${\log }_3\sqrt[3]{9}={\log }_3\sqrt[3]{3^2}={\log }_3{3}^{\frac{2}{3}}$ $=\frac{2}{3}{\log }_{3}3=\frac{2}{3}.1=\frac{2}{3}.$
c) $9^{{\log }_{3}12}={\left(3^2\right)}^{{\log }_{3}12}={\left(3^{{\log }_{3}12}\right)}^2={12}^2=144.$
d) $2^{{\log }_{4}9}={\left(2\right)}^{{\log }_{2^2}9}=2^{\frac{1}{2}{\log }_{2}9}={\left(2^{{\log }_{2}9}\right)}^{\frac{1}{2}}=9^{\frac{1}{2}}=3.$

a) $A=\frac{{25}^{{\log }_{5}6}+{49}^{{\log }_{7}8}-3}{3^{1+{\log }_{9}4}+4^{2-{\log }_{2}3}+5^{{\log }_{125}27}}$
$=\frac{{\left(5^2\right)}^{{\log }_{5}6}+{\left(7^2\right)}^{{\log }_{7}8}-3}{3\cdot 3^{{\log }_{9}4}+\frac{4^2}{4^{{\log }_{2}3}}+5^{{\log }_{5^3}\left(3^3\right)}}$$=\frac{{\left(5^{{\log }_{5}6}\right)}^2+{\left(7^{{\log }_{7}8}\right)}^2-3}{3\cdot 3^{{\log }_{3^2}\left(2^2\right)}+\frac{4^2}{{\left(2^2\right)}^{{\log }_{2}3}}+5^{\frac{1}{3}\cdot 3\cdot {\log }_{5}3}}$

$=\frac{6^2+8^2-3}{3\cdot 3^{{\log }_{3}2}+\frac{4^2}{{\left(2^{{\log }_{2}3}\right)}^2}+5^{{\log }_{5}3}}$$=\frac{36+64-3}{3\cdot 2+\frac{4^2}{3^2}+3}=\frac{97}{6+\frac{16}{9}+3}=\frac{97}{\frac{97}{9}}=9.$

b) $B=\frac{{36}^{{\log }_{6}5}+{10}^{1-\log 2}-3^{{\log }_{9}36}}{{\log }_2\left({\log }_2\sqrt{\sqrt[4]{2}}\right)}$
$=\frac{{\left(6^2\right)}^{{\log }_{6}5}+\frac{10}{{10}^{\log 2}}-3^{{\log }_{3^2}\left(6^2\right)}}{{\log }_2\left({\log }_2\sqrt{2^{\frac{1}{4}}}\right)}$$=\frac{{\left(6^{{\log }_{6}5}\right)}^2+\frac{10}{2}-3^{\frac{1}{2}\cdot 2\cdot {\log }_{3}6}}{{\log }_2\left({\log }_2{2}^{\frac{1}{8}}\right)}$
$=\frac{5^2+5-3^{{\log }_{3}6}}{{\log }_2\left(\frac{1}{8}\right)}$$=\frac{25+5-6}{{\log }_2{2}^{-3}}=\frac{24}{-3}=-8.$
c) $C={\log }_{\frac{1}{4}}\left({\log }_{3}4.{\log }_{2}3\right)$
$={\log }_{2^{-2}}\left({\log }_3{2}^2\cdot {\log }_{2}3\right)$$=-\frac{1}{2}{\log }_2\left(2{\log }_{3}2\cdot {\log }_{2}3\right)$$=-\frac{1}{2}{\log }_2\left(2{\log }_{3}2\cdot \frac{1}{{\log }_{3}2}\right)$$=-\frac{1}{2}{\log }_{2}2=-\frac{1}{2}.$
d) D = log₄ 2 . log₆ 4 . log₈ 6
$={\log }_{4}2\cdot \frac{{\log }_{4}4}{{\log }_{4}6}\cdot \frac{{\log }_{4}6}{{\log }_{4}8}$$={\log }_{4}2\cdot \frac{{\log }_{4}4}{{\log }_{4}8}$
$={\log }_{4}2\cdot {\log }_{8}4={\log }_{8}4\cdot {\log }_{4}2$
$={\log }_{8}2={\log }_{2^3}2=\frac{1}{3}{\log }_{2}2=\frac{1}{3}.$

a) ${\log }_a\left(a^{\frac{1}{2}}{b}^5\right)={\log }_a{a}^{\frac{1}{2}}+{\log }_a{b}^5$
$=\frac{1}{2}.{\log }_{a}a+5.{\log }_{a}b=\frac{1}{2}.1+5.4=\frac{41}{2}.$
b) ${\log }_a\left(\frac{a\sqrt{b}}{b\sqrt[3]{a}}\right)={\log }_a\left(\frac{a.b^{\frac{1}{2}}}{a^{\frac{1}{3}}b}\right)$
$={\log }_a\left(\frac{a}{a^{\frac{1}{3}}}.\frac{b^{\frac{1}{2}}}{b}\right)={\log }_a\left(a^{1-\frac{1}{3}}.b^{\frac{1}{2}-1}\right)$
$={\log }_a\left(a^{\frac{2}{3}}{b}^{-\frac{1}{2}}\right)={\log }_a{a}^{\frac{2}{3}}+{\log }_a{b}^{-\frac{1}{2}}$
$=\frac{2}{3}{\log }_{a}a-\frac{1}{2}{\log }_{a}b=\frac{2}{3}.1-\frac{1}{2}.4=-\frac{4}{3}.$
c) ${\log }_{a^3{b}^2}\left(a^2{b}^3\right)=\frac{{\log }_a\left(a^2{b}^3\right)}{{\log }_a{a}^3{b}^2}$
$=\frac{{\log }_a{a}^2+{\log }_a{b}^3}{{\log }_a{a}^3+{\log }_a{b}^2}$
$=\frac{2+3{\log }_{a}b}{3+2{\log }_{a}b}$$=\frac{2+3.4}{3+2.4}=\frac{14}{11}.$
d) ${\log }_{a\sqrt[3]{b}}\left(\sqrt[4]{a\sqrt{b}}\right)=\frac{{\log }_a\sqrt[4]{a\sqrt{b}}}{{\log }_{a}a\sqrt[3]{b}}$
$=\frac{{\log }_a\sqrt[4]{a{b}^{\frac{1}{2}}}}{{\log }_a\left(a{b}^{\frac{1}{3}}\right)}=\frac{{\log }_a{\left(a{b}^{\frac{1}{2}}\right)}^{\frac{1}{4}}}{{\log }_a\left(a{b}^{\frac{1}{3}}\right)}$ 

a) ${\log }_{18}72=\frac{{\log }_{2}72}{{\log }_{2}18}=\frac{{\log }_2\left(2^3{.3}^2\right)}{{\log }_2\left({2.3}^2\right)}$$=\frac{{\log }_2{2}^3+{\log }_2{3}^2}{{\log }_{2}2+{\log }_2{3}^2}=\frac{3+2{\log }_{2}3}{1+2{\log }_{2}3}=\frac{3+2a}{1+2a}.$

b*)  ${\log }_{20}50=\frac{\log 50}{\log 20}=\frac{\log \left(10.5\right)}{\log \left(10.2\right)}$$=\frac{\log 10+\log 5}{\log 10+\log 2}=\frac{1+1-a}{1+a}=\frac{2-a}{1+a}.$

Với x > 0, y > 0 ta có:

x² + 4y² = 6xy ⇒ x² + 4xy + 4y² = 10xy

                   ⇒ (x + 2y)² = 10xy.

Suy ra: 2log(x + 2y) = log(x + 2y)²

        = log(10xy) = log10 + logx + logy

        = 1 + logx + logy.

Vậy 2log(x + 2y) = 1 + logx + logy.

Do log_xa, log_yb, log_zc theo thứ tự lập thành một cấp số cộng nên ta có: 

$2{\log }_{y}b={\log }_{x}a+{\log }_{z}c$

$\iff 2{\log }_{y}b=\frac{1}{{\log }_{a}x}+\frac{1}{{\log }_{c}z}$

$\iff 2{\log }_{y}b=\frac{{\log }_{c}z+{\log }_{a}x}{{\log }_{a}x.{\log }_{c}z}$

$\iff \frac{2}{{\log }_{b}y}=\frac{{\log }_{a}x+{\log }_{c}z}{{\log }_{a}x.{\log }_{c}z}$

$\Rightarrow {\log }_{b}y=\frac{2{\log }_{a}x.{\log }_{c}z}{{\log }_{a}x+{\log }_{c}z}$

Chất phóng xạ có chu kì bán rã là T = 5 730 (năm).

Suy ra:  $\lambda =\frac{\ln 2}{5730}.$

Gọi t là độ tuổi của mẫu gỗ cổ.

Vì độ phóng xạ của mẫu gỗ tươi cùng loại là 0,250 Bq nên ta có H₀ = 0,250 Bq.

Khi khảo sát một mẫu gỗ cổ, các nhà khoa học đo được độ phóng xa là 0,215 Bq, suy ra ta có H = 0,215 Bq.

Ta có: $H=H_0{e}^{-\lambda t}\Rightarrow 0,215=0,250\cdot e^{-\frac{\ln 2}{5730}\cdot t}$

$\Rightarrow e^{-\frac{\ln 2}{5730}\cdot t}=\frac{0,215}{0,250}=\frac{43}{50}$

$\Rightarrow e^{\frac{\ln 2}{5730}\cdot t}=\frac{50}{43}\Rightarrow \frac{\ln 2}{5730}\cdot t=\ln \frac{50}{43}$

$\Rightarrow t=\ln \frac{50}{43}:\frac{\ln 2}{5730}\approx 1247.$

Vậy độ tuổi của mẫu gỗ cổ đó xấp xỉ 1 247 năm.