تمارين (1-1)

(1)- جد قيمة × لكل مما يأتي:

أ- ${\log }_{10}0.00001=\mathrm{X}$

$\frac{0.00001}{1}={10}^{\mathrm{x}}\to \frac{1}{100000}={10}^{\mathrm{x}}\to {10}^{-5}={10}^{\mathrm{x}}\to \mathrm{x}=-5$

ب- ${\log }_{\mathrm{x}}16=-4$

$16={\mathrm{x}}^{-4}\to 2^4=\frac{1}{{\mathrm{x}}^4}\to 2^4\cdot {\mathrm{x}}^4=1\to {\mathrm{x}}^4=\frac{1}{2^4}\to {\mathrm{x}}^4=2^{-4}{\mathrm{x}}^4={\left(2^{-1}\right)}^4\to \mathrm{x}=\mp 2^{-1}\to \mathrm{x}=\mp \frac{1}{2}\to \mathrm{x}=\frac{1}{2}$

ج- ${\log }_{10}\mathrm{x}=5$

$X={10}^5\to X=100000$

(2)- اكتب الصورة الأخرى لكل مما يأتي:

أ- ${\log }_{10}10000=4$

$10000={10}^4$

ب- $7^3=343$

${\log }_{7}343=3$

ج- ${\log }_5\frac{1}{25}=-2$

$\frac{1}{25}=5^{-2}$

د- $(0.01{)}^2=0.0001$

${\log }_{0.01}0.0001=2$

(3)- فيما يلي علاقات غير صحيحة دائماً، أعط y = a ، x = a ، حيث 0 < a وبين ذلك:

أ- ${\log }_{\mathrm{a}}(\mathrm{x}+\mathrm{y})\ne {\log }_2\mathrm{x}+{\log }_{\mathrm{a}}\mathrm{y}$

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

ب- ${\log }_{\mathrm{a}}\mathrm{xy}\ne {\log }_{\mathrm{a}}\mathrm{x}.{\log }_{\mathrm{a}}\mathrm{y}$

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

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

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

(4)- جد قيمة ما يأتي:

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

$\begin{array}{rl}& \to {\log }_{10}\frac{40}{9}+{\log }_{12}{5}^4+{\log }_{10}{6}^2\\ & \to {\log }_{10}\frac{40}{9}\cdot 5^4\cdot 6^2\\ & \to {\log }_{10}\frac{40.625\cdot 36}{9}\to {\log }_{10}100000\to {\log }_{10}{10}^5\to 5{\log }_{10}10\\ & \to 5.1=5\end{array}$

ب- $2{\log }_{10}8+{\log }_{10}125-3{\log }_{10}20$

$\begin{array}{rl}& ={\log }_{10}{8}^2+{\log }_{10}125-{\log }_{10}{20}^3\\ & ={\log }_{10}64+{\log }_{10}125-{\log }_{10}\left(\mathrm{20.20.20}\right)\\ & ={\log }_{10}\frac{64.125}{\mathrm{20.20.20}}\to {\log }_{10}\frac{8000}{8000}\\ & ={\log }_{10}1=0({\mathrm{a}}^0=1)\end{array}$

ج- ${\log }_{\mathrm{a}}({\mathrm{x}}^2-4)-2{\log }_{\mathrm{a}}(\mathrm{x}-2)+{\log }_{\mathrm{a}}\frac{\mathrm{x}-2}{\mathrm{x}+2}$

$\begin{array}{rl}& \to {\log }_{\mathrm{a}}({\mathrm{x}}^2-4)-{\log }_{\mathrm{a}}(\mathrm{x}-2{)}^2+{\log }_{\mathrm{a}}\frac{\mathrm{x}-2}{\mathrm{x}+2}\\ & \to {\log }_{\mathrm{a}}\frac{({\mathrm{x}}^2-4)(\mathrm{x}-2)}{(\mathrm{x}-2{)}^2(\mathrm{x}+2)}\\ & \to {\log }_{\mathrm{a}}\frac{(\mathrm{x}-2)\mathrm{x}+2\left)\right(\mathrm{x}-2)}{(\mathrm{x}-2)(\mathrm{x}-2)(\mathrm{x}+2)}\to {\log }_{\mathrm{a}}1\\ & ={\log }_{\mathrm{a}}1=0({\mathrm{a}}^0=1)\end{array}$

(5)- إذا كان ${\log }_{10}3=0.4771,{\log }_{10}2=0.3010$ جد قيمة كل مما يأتي:

أ- ${\log }_{10}\frac{0.002}{1}={\log }_{10}\frac{2}{1000}$

$\begin{array}{rl}& {\log }_{10}\frac{0.002}{1}={\log }_{10}\frac{2}{1000}\\ & \to {\log }_{10}2-{\log }_{10}1000\to 0.3010-3\to -{3.3010}^{}\end{array}$

ب- ${\log }_{10}2000$

$\begin{array}{rl}& {\log }_{10}2000\to {\log }_{10}2\times 1000\to {\log }_{10}2+{\log }_{10}1000\\ & \to {\log }_{10}2+{\log }_{10}{10}^3\to {\log }_{10}2+3{\log }_{10}10\to {\log }_{10}2+(3\times 1)\\ & \to 0.3010+3\to 3.3010\end{array}$

ج- ${\log }_{10}12$

$\begin{array}{rl}{\log }_{10}12={\log }_{10}3.4& ={\log }_{10}3+{\log }_{10}4\\ & ={\log }_{10}3+{\log }_{10}{2}^2\\ & ={\log }_{10}3+2{\log }_{10}2\\ & =0.4771+2\left(0.3010\right)\\ & =0.4771+0.6020=1.0791\end{array}$

(6)- حل المعادلات الآتية:

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

$\begin{array}{rl}& {\log }_2(2x-1)(x+4)={\log }_{3}5\\ & (2x-1)(x+4)=5\\ & 2x(x+4)-1(x+4)=5\\ & 2x^2+8x-x-4=5\\ & 2x^2+7x-4-5=0\\ & 2x^2+7x-9=0\\ & (2x+9)(x+1)\end{array}$

نضرب الطرفين بالوسطين

$9x-2x=7x$

إما: $(\mathrm{x}-1)=0\to \mathrm{x}=1$

أو: $(2x+9)=0\to 2x=-9\to x=\frac{-9}{2}$

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

نحول إلى دالة أسية

$\begin{array}{rl}& {\log }_2\frac{3\mathrm{x}+5}{\mathrm{x}-5}=3\\ & \frac{3\mathrm{x}+5}{\mathrm{x}-5}=2^3\to \frac{3\mathrm{x}+5}{\mathrm{x}-5}=8\to 5\mathrm{x}-40=3\mathrm{x}+5\\ & \to 8\mathrm{x}-3\mathrm{x}=40+5\to 5\mathrm{x}=\frac{5\mathrm{x}}{5}=\frac{45}{5}\\ & \to \mathrm{x}=9\end{array}$

ج- ${\log }_{\mathrm{a}}\frac{6}{5}+{\log }_{\mathrm{a}}\frac{5}{66}-{\log }_{\mathrm{a}}\frac{132}{121}+{\log }_{\mathrm{a}}12=\mathrm{x}$

نحول إلى دالة أسية

$\begin{array}{rl}& {\log }_{\mathrm{a}}(\frac{6}{5}\cdot \frac{5}{66}\cdot \frac{12}{1})÷\frac{132}{121}=\mathrm{X}\\ & {\log }_{\mathrm{a}}\frac{12}{11}\times \frac{121}{132}=\mathrm{X}\to {\log }_{\mathrm{a}}\frac{12}{11}\times \frac{11\times 11}{11\times 12}=\mathrm{X}\\ & {\log }_{\mathrm{a}}1=\mathrm{X}\\ & 1=\mathrm{a}\\ & 1={\mathrm{a}}^0={\mathrm{a}}^{\mathrm{X}}\\ & \mathrm{X}=0\end{array}$

د- ${\log }_{10}(3\mathrm{x}-7)+{\log }_{10}(3\mathrm{x}+1)=1+{\log }_{10}2$

$\begin{array}{rl}& {\log }_{10}\mathrm{Z}=1\to \mathrm{Z}={10}^1\to \mathrm{Z}=10\\ & {\log }_{10}(3\mathrm{x}-7)+{\log }_{10}(3\mathrm{x}+1)={\log }_{10}10+{\log }_{10}2\\ & {\log }_{10}(3\mathrm{x}-7)(3\mathrm{x}+1)={\log }_{10}10\times 2\\ & (3\mathrm{x}-7)(3\mathrm{x}+1)=10\times 2\\ & \begin{array}{rl}& 3\mathrm{x}(3\mathrm{x}+1)-7(3\mathrm{x}+1)=20\\ & 9{\mathrm{x}}^2+3\mathrm{x}-21\mathrm{x}-7-20=0\\ & (9{\mathrm{x}}^2-18\mathrm{x}-27=0)\times \frac{1}{9}\\ & {\mathrm{x}}^2-2\mathrm{x}-3=0\\ & (\mathrm{x}-3)(\mathrm{x}+1)=0\end{array}\end{array}$

إما: $(\mathrm{x}-3)=0\to \mathrm{x}=3$

أو: $\mathrm{x}+1=0\to \mathrm{x}=-1$