جواب تمرین صفحه 113 درس 5 ریاضی یازدهم تجربی (توابع نمایی و لگاریتمی)

الف

\(\begin{array}{l}{\log _c}abd = {\log _c}a + {\log _c}b + {\log _c}d\\\\ \Rightarrow {\log _c}abd = {\log _c}\left( {ab} \right)d = {\log _c}ab + {\log _c}d\\\\ = {\log _c}a + {\log _c}b + {\log _c}d\end{array}\)

ب

\(\begin{array}{l}{\log _b}a = \frac{{{{\log }_c}a}}{{{{\log }_c}b}}\\\\ \Rightarrow {\log _b}a = x \Rightarrow a = {b^x}\mathop \to \limits^{{{\log }_c}\left( {} \right)} {\log _c}a = {\log _c}{b^x}\\\\ \Rightarrow {\log _c}a = x{\log _c}b\\\\ \Rightarrow x = \frac{{{{\log }_c}a}}{{{{\log }_c}b}} \Rightarrow {\log _b}a = \frac{{{{\log }_c}a}}{{{{\log }_c}b}}\end{array}\)

پ

\(\begin{array}{l}{a^{{{\log }_a}b}} = b\\\\\mathop \to \limits^{{{\log }_a}\left( {} \right)} {\log _a}{a^{{{\log }_a}b}} = {\log _a}b \Rightarrow {\log _a}b \times {\not{{{{\log }_a}a}}^1}\\\\ = {\log _a}b \Rightarrow {\log _a}b = {\log _a}b\end{array}\)

ت

\(\begin{array}{l}{\log _b}a \times {\log _a}b = 1\\\\ \Rightarrow \left. \begin{array}{l}{\log _b}a = \frac{{{{\log }_c}a}}{{{{\log }_c}b}}\\{\log _a}b = \frac{{{{\log }_c}b}}{{{{\log }_c}a}}\end{array} \right\}\\\\ \Rightarrow {\log _b}a \times {\log _a}b = \frac{{\not{{{{\log }_c}a}}}}{{\not{{{{\log }_c}b}}}} \times \frac{{\not{{{{\log }_c}b}}}}{{\not{{{{\log }_c}a}}}} = 1\end{array}\)

الف

\({\log _7}\sqrt[5]{{49}} = {\log _7}{\left( {49} \right)^{\frac{1}{5}}} = {\log _7}{\left( {{7^2}} \right)^{\frac{1}{5}}} = {\log _7}{7^{\frac{2}{5}}} = \frac{2}{5}{\log _7}7 = \frac{2}{5}\)

ب

\({\log _3}{27^{\frac{1}{2}}} = {\log _3}{\left( {{3^3}} \right)^{\frac{1}{2}}} = {\log _3}{3^{\frac{3}{2}}} = \frac{3}{2}{\log _3}2 = \frac{3}{2}\)

پ

\( - {\log _5}125 = - {\log _5}{5^3} = - 3{\log _5}5 = - 3\)

ت

\(\begin{array}{l}3{\log _{10}}\sqrt {1000} = 3{\log _{10}}{\left( {1000} \right)^{\frac{1}{2}}} = 3{\log _{10}}{\left( {{{10}^3}} \right)^{\frac{1}{2}}} = 3{\log _{10}}{10^{\frac{3}{2}}}\\\\ = 3 \times \frac{3}{2}{\log _{10}}10 = \frac{9}{2}\end{array}\)

\(\begin{array}{l}f\left( x \right) = 3 - 2{\log _4}(\frac{x}{2} - 5) \Rightarrow f\left( {42} \right) = 3 - 2{\log _4}(\frac{{42}}{2} - 5)\\\\ = 3 - 2{\log _4}16 = 3 - 2 \times 2 = - 1\end{array}\)

الف

\(\left. \begin{array}{l}f\left( x \right) = {\log _a}x\\\left( {2\;,\;2} \right) \in f\end{array} \right\} \Rightarrow 2 = {\log _a}2 \Rightarrow {a^2} = 2\;\;\mathop \Rightarrow \limits^{a > 0} \;\;a = \sqrt 2 \)

ب

\(\begin{array}{l}\left. \begin{array}{l}f\left( x \right) = {\log _a}x\\(\frac{1}{2}\;,\; - 4) \in f\end{array} \right\} \Rightarrow - 4 = {\log _a}\frac{1}{2}\\\\ \Rightarrow {a^{ - 4}} = \frac{1}{2} \Rightarrow {a^4} = 2\;\mathop \Rightarrow \limits^{a > 0} \;a = \sqrt[4]{2}\end{array}\)

الف

درست نیست.

ب

درست است.

پ

درست است.

الف

\(\begin{array}{l}{\log _3}\left( {{p^2} - 2} \right) = {\log _3}p\\\\ \Rightarrow {p^2} - 2 = p \Rightarrow {p^2} - p - 2 = 0\\\\ \Rightarrow \left( {p + 1} \right)\left( {p - 2} \right) = 0 \Rightarrow \left\{ \begin{array}{l}{p_1} = - 1\not|{{}}\\{p_2} = 2\end{array} \right.\;\mathop \Rightarrow \limits^{p > 0} \;p = 2\end{array}\)

ب

\(\begin{array}{l}{\log _5}\left( {x + 1} \right) + {\log _5}\left( {x - 1} \right) = 1\\\\ \Rightarrow {\log _5}\left( {x + 1} \right)\left( {x - 1} \right) = 1 \Rightarrow {\log _5}\left( {{x^2} - 1} \right) = 1\\\\ \Rightarrow {x^2} - 1 = 5 \Rightarrow {x^2} = 6 \Rightarrow x = \pm \sqrt 6 \\\\\mathop \Rightarrow \limits^{x > 1} \;x = \sqrt 6 \end{array}\)

پ

\(\begin{array}{l}3{\log _4}a - {\log _4}5 = {\log _4}25\\\\ \Rightarrow {\log _4}{a^3} - {\log _4}5 = {\log _4}\frac{{{a^3}}}{5}\\\\ = {\log _4}25 \Rightarrow \frac{{{a^3}}}{5} = 25 \Rightarrow {a^3} = 125\\\\ \Rightarrow a = \sqrt[3]{{125}} = 5\end{array}\)

ت

\(\begin{array}{l}{\log _{\frac{1}{{10}}}}\left( {{x^2} - 21} \right) = - 2\\\\ \Rightarrow {x^2} - 21 = {(\frac{1}{{10}})^{ - 2}} = {({10^{ - 1}})^{ - 2}} = {10^2} = 100\\\\ \Rightarrow {x^2} = 121 \Rightarrow x = \pm \sqrt {121} \Rightarrow \left\{ \begin{array}{l}{x_1} = - 11\\{x_2} = 11\end{array} \right.\end{array}\)