تمارين (1-1)

(1)- فيما يلي علاقات غير صحيحة دائماً، مثلاً أعط x=a، y=a وبين ذلك.

أ- ${\log }_{\mathrm{a}}(\mathrm{x}+\mathrm{y})\ne {\log }_{\mathrm{a}}\mathrm{x}+{\log }_{\mathrm{a}}\mathrm{y}$

$\begin{array}{rl}& {\log }_{\mathrm{a}}(\mathrm{x}+\mathrm{y})\ne {\log }_{\mathrm{a}}\mathrm{x}+{\log }_{\mathrm{a}}\mathrm{y}\\ & {\log }_{\mathrm{a}}(\mathrm{a}+\mathrm{a})\ne {\log }_{\mathrm{a}}\mathrm{a}+{\log }_{\mathrm{a}}\mathrm{a}\\ & {\log }_{\mathrm{a}}(\mathrm{a}+\mathrm{a})\ne {\log }_{\mathrm{a}}\mathrm{a}+{\log }_{\mathrm{a}}\mathrm{a}\\ & {\log }_{\mathrm{a}}\left(2\mathrm{a}\right)\ne {\log }_{\mathrm{a}}\mathrm{a}+{\log }_{\mathrm{a}}\mathrm{a}\\ & {\log }_{\mathrm{a}}2+{\log }_{\mathrm{a}}\mathrm{a}\ne {\log }_{\mathrm{a}}\mathrm{a}+{\log }_{\mathrm{a}}\mathrm{a}\\ & {\log }_{\mathrm{a}}2+1\ne 1+1\\ & 0.3010+1\ne 2\\ & 1.3010\ne 2\end{array}$

ب- ${\log }_{\mathrm{a}}(\mathrm{x}-\mathrm{y})\ne \frac{{\log }_{\mathrm{a}}\mathrm{x}}{{\log }_{\mathrm{a}}\mathrm{y}}$

$\begin{array}{rl}& {\log }_{\mathrm{a}}(\mathrm{x}-\mathrm{y})\ne \frac{{\log }_{\mathrm{a}}\mathrm{x}}{{\log }_{\mathrm{a}}\mathrm{y}}\\ & {\log }_{\mathrm{a}}(\mathrm{x}-\mathrm{y})\ne {\log }_{\mathrm{a}}\mathrm{x}-{\log }_{\mathrm{a}}\mathrm{y}\\ & {\log }_{\mathrm{a}}(\mathrm{a}-\mathrm{a})\ne {\log }_{\mathrm{a}}\mathrm{a}-{\log }_{\mathrm{a}}\mathrm{a}\\ & {\log }_{\mathrm{a}}\left(0\right)\ne 1-1\\ & {\log }_{\mathrm{a}}\left(0\right)\ne 0\end{array}$

ج- ${\log }_{\mathrm{a}}\mathrm{xy}\ne {\log }_{\mathrm{a}}{\mathrm{xlog}}_{\mathrm{a}}\mathrm{y}$

$\begin{array}{rl}& {\log }_{\mathrm{a}}(\mathrm{a},\mathrm{a})\ne {\log }_{\mathrm{a}}\mathrm{a}\cdot {\log }_{\mathrm{a}}\mathrm{a}\\ & {\log }_{\mathrm{a}}{\mathrm{a}}^2\ne {\log }_{\mathrm{a}}\mathrm{a}\cdot {\log }_{\mathrm{a}}\mathrm{a}\\ & 2{\log }_{\mathrm{a}}\mathrm{a}\ne {\log }_{\mathrm{a}}\mathrm{a}\cdot {\log }_{\mathrm{a}}\mathrm{a}\\ & 2\times 1\ne 1\times 1\\ & 2\ne 1\end{array}$

د- ${\log }_{\mathrm{a}}{\mathrm{x}}^2\ne ({\log }_{\mathrm{a}}{\mathrm{x}}^2$

$\begin{array}{rl}& {\log }_a{a}^2\ne {({\log }_{a}a)}^2\\ & 2{\log }_{a}a\ne {({\log }_{a}a)}^2\\ & 2\times 1\ne (1{)}^2\\ & 2\ne 1\end{array}$

(2)- جد قيمة x:

أ- ${\log }_{10}0.001=\mathrm{x}$

$$\begin{gathered} {\log }_{10}0.001=\mathrm{x}\to {10}^{\mathrm{x}}=0.001 \\ {10}^{\mathrm{x}}=\frac{1}{{10}^3}\to {10}^{\mathrm{x}}={10}^{-3}\to \mathrm{x}=-3 \end{gathered}$$

ب- ${\log }_{\mathrm{x}}\frac{1}{8}=-3$

$\begin{array}{rl}& {\log }_{\mathrm{x}}\frac{1}{8}=-3\to {\mathrm{x}}^{-3}=\frac{1}{8}\\ & \frac{1}{{\mathrm{x}}^3}=\frac{1}{2^3}\to {\mathrm{x}}^3=2^3\to \mathrm{x}=2\end{array}$

ج- ${\log }_{10}\mathrm{x}=5$

$\begin{array}{rl}& {\log }_{10}\mathrm{x}=5\to \mathrm{x}={10}^5\\ & \mathrm{x}=100000\end{array}$

(3)- جد قيمة ما يأتي:

أ- ${\log }_{10}\left(\frac{40}{9}\right)+4{\log }_{10}5+2{\log }_{10}6$

$\begin{array}{rl}& {\log }_{40}\frac{40}{9}+4{\log }_{10}5+2{\log }_{10}6\\ & ={\log }_{10}\frac{40}{9}+{\log }_{10}(5{)}^4+{\log }_{10}(6{)}^2\\ & ={\log }_{10}\frac{40}{9}\times (5\times 5\times 5\times 5)\times (6\times 6)\\ & ={\log }_{10}\frac{40}{9}\times (25\times 25)\times \left(36\right)\\ & ={\log }_{10}\frac{40}{9}\times \frac{625}{1}\times \frac{36}{1}\\ & ={\log }_{10}40\times 625\times 4\\ & ={\log }_{10}100000={\log }_{10}{10}^5\\ & =5{\log }_{10}10=5\times 1=5\end{array}$

ب- $2{\log }_{10}8+{\log }_{10}125-3{\log }_{10}200$

$\begin{array}{rl}& 2{\log }_{50}8+{\log }_{10}125-3{\log }_{10}200\\ & ={\log }_{10}(8{)}^2+{\log }_{10}125-3{\log }_{10}(200{)}^3\\ & ={\log }_{1.2}\frac{64\times 125}{200\times 200\times 200}\\ & ={\log }_{10}\frac{8000}{8000000}={\log }_{10}\frac{1}{1000}\\ & ={\log }_{10}\frac{1}{1000}={\log }_{10}\frac{1}{{10}^3}={\log }_{10}{10}^{-3}\\ & =-3{\log }_{10}10=-3\times 1=-3\end{array}$

ج- ${\log }_{\mathrm{a}}({\mathrm{x}}^2-1)-2{\log }_{\mathrm{a}}(\mathrm{x}-1)+{\log }_{\mathrm{a}}\frac{(\mathrm{x}-1)}{(\mathrm{x}+1)}$

$\begin{array}{rl}& {\log }_{\mathrm{a}}({\mathrm{x}}^2-1)-2{\log }_{\mathrm{a}}(\mathrm{x}-1)+{\log }_{\mathrm{a}}\frac{(\mathrm{x}-1)}{(\mathrm{x}+1)}\\ & ={\log }_{\mathrm{a}}({\mathrm{x}}^2-1)-{\log }_{\mathrm{a}}(\mathrm{x}-1{)}^2+{\log }_{\mathrm{a}}\frac{(\mathrm{x}-1)}{(\mathrm{x}+1)}\\ & ={\log }_{\mathrm{a}}\frac{({\mathrm{x}}^2-1)}{(\mathrm{x}-1{)}^2}\times \frac{(\mathrm{x}-1)}{(\mathrm{x}+1)}={\log }_{\mathrm{a}}\frac{(\mathrm{x}-1)(\mathrm{x}+1)}{(\mathrm{x}+1)(\mathrm{x}-1)}\times \frac{(\mathrm{x}-1)}{(\mathrm{x}+1)}={\log }_{\mathrm{a}}1=0\end{array}$

د- ${\log }_{2}8-{\log }_{3}27-{\log }_{5}625$

$\begin{array}{rl}& {\log }_{2}8-{\log }_{3}27-{\log }_{5}625={\log }_2{2}^3-{\log }_3{3}^3-{\log }_5{5}^4\\ & =3{\log }_{2}2-3{\log }_{3}3-4{\log }_{5}5=3\times 1-3\times 1-4\times 1\\ & =3-3-4=-4\end{array}$

(4)- إذا كانت ${\log }_{10}2=0.3010 , {\log }_{10}3=0.4771$ جد قيمة:

أ- ${\log }_{10}0.002$

$\begin{array}{rl}& ={\log }_{10}\frac{2}{1000}\\ & ={\log }_{10}2-{\log }_{10}1000\\ & ={\log }_{10}2-{\log }_{48}{10}^3\\ & ={\log }_{10}2-3{\log }_{50}10\\ & =0.3010-(3\times 1)\\ & =0.3010-3\\ & =-3.3010\end{array}$

ب- ${\log }_{10}3000$

$\begin{array}{rl}& {\log }_{10}3000\\ & ={\log }_{10}3\times 1000\\ & ={\log }_{10}3+{\log }_{10}{10}^3\\ & ={\log }_{10}3-3{\log }_{10}10\\ & =0.4771+(3\times 1)\\ & =0.4771+3\\ & =3.4771\end{array}$

ج- ${\log }_{10}12$

$\begin{array}{rl}& {\log }_{10}12\\ & ={\log }_{10}3\times 4\\ & ={\log }_{10}3\times 2\times 2\\ & ={\log }_{10}3+{\log }_{10}2+{\log }_{10}2\\ & =0.4771+0.3010+0.3010\\ & =1.0791\end{array}$

(5)- حل المعادلات الآتية:

أ- ${\log }_3(2\mathrm{x}-1)+{\log }_3(\mathrm{x}+4)={\log }_{3}5$

نشاهد الأساسات متساوية فنستخدم قانوني الجمع والطرح في اللوغاريتمات:

$\begin{array}{rl}& {\log }_3(2\mathrm{x}-1)+{\log }_3(\mathrm{x}+4)={\log }_{3}5\\ & {\log }_3(2\mathrm{x}-1)\times (\mathrm{x}+4)={\log }_{3}5\\ & {\log }_3(2\mathrm{x}-1)\times (\mathrm{x}+4)={\log }_{3}5\\ & (2\mathrm{x}-1)(\mathrm{x}+4)=5\\ & (2\mathrm{x}-1)(\mathrm{x}+4)-5=5-5\\ & 2{\mathrm{x}}^2+8\mathrm{x}-\mathrm{x}-4-5=0\\ & 2{\mathrm{x}}^2+7\mathrm{x}-9=0\\ & (2\mathrm{x}+9)(\mathrm{x}-1)=0\\ & 2\mathrm{x}+9=0\Rightarrow 2\mathrm{x}=-9\to \mathrm{x}=\frac{-9}{2}\to \notin {\mathrm{R}}^{+}\\ & \mathrm{x}-1=0\to \mathrm{x}=1\\ & \mathrm{S}=\left\{1\right\}\end{array}$

ب- ${\log }_2(3\mathrm{x}+5)-{\log }_2(\mathrm{x}-5)=3$

نشاهد الأساسات متساوية فنستخدم قانوني الجمع والطرح في اللوغاريتمات:

$\begin{array}{rl}& {\log }_2(3\mathrm{x}+5)-{\log }_2(\mathrm{x}-5)=3\\ & {\log }_3\frac{(3\mathrm{x}+5)}{(\mathrm{x}-5)}=3\leftrightarrow \frac{(3\mathrm{x}+5)}{(\mathrm{x}-5)}=2^3\end{array}$

حولنا المعادلة من صورة لوغاريتمية إلى صورة أسية:

$$\begin{gathered} {\mathrm{a}}^{\mathrm{x}}=\mathrm{y} \\ \begin{array}{rl}& \frac{(3\mathrm{x}+5)}{(\mathrm{x}-5)}=\frac{8}{1}\\ & 8(\mathrm{x}-5)=(3\mathrm{x}+5)\\ & 8\mathrm{x}-8\times 5=3\mathrm{x}+5\\ & 8\mathrm{x}-3\mathrm{x}-40=3\mathrm{x}-3\mathrm{x}+5\\ & 5\mathrm{x}-40-5=5-5\\ & 5\mathrm{x}-45=0\to 5\mathrm{x}=45\to \mathrm{x}=\frac{45}{5}=9\in {\mathrm{R}}^{+}\\ & \mathrm{S}=\left\{9\right\}\end{array} \end{gathered}$$