جواب تمرین صفحه 90 درس 3 حسابان یازدهم

الف

\(\begin{array}{l}{\log _4}{m^2} - {\log _4}m - 3 = 0 \Rightarrow {\log _4}\left( {\frac{{{m^2}}}{{3m}}} \right) = 0\\\\ \Rightarrow \frac{m}{3} = 1 \Rightarrow m = 1\end{array}\)

ب

\(\begin{array}{l}{\log _2}\left( {12b - 21} \right) - {\log _2}\left( {{b^2} - 3} \right) = 2 \Rightarrow {\log _2}\left\{ {\frac{{12b - 21}}{{{b^2} - 3}}} \right\} = 2\\\\ \Rightarrow \frac{{12b - 21}}{{{b^2} - 3}} = 4\\\\ \Rightarrow 4{b^2} - 12 = 12b - 21 \Rightarrow 4{b^2} - 12b + 9 = 0\\\\ \Rightarrow {\left( {2b - 3} \right)^2} = 0 \Rightarrow b = \frac{3}{2}\\\\ \Rightarrow 12b - 21 = 18 - 21 = - 3\end{array}\)

پ

\(\begin{array}{l}{\log _{\frac{1}{{10}}}}\left( {{x^2} - 1} \right) = - 1 \Rightarrow {x^2} - 1 = {\left( {\frac{1}{{10}}} \right)^{ - 1}}\\\\ \Rightarrow {x^2} - 1 = 10\\\\ \Rightarrow {x^2} = 9 \Rightarrow \left\{ \begin{array}{l}x = 3\\x = - 3\end{array} \right.\end{array}\)

الف

\(m\left( t \right) = {2^t} \Leftrightarrow {m^{ - 1}}\left( t \right) = {\log _2}m\left( t \right) = t\)

تفسیرش این است که اگر بخواهیم بدانیم که چه میزان زمان لازم است تا (t)m گرم باکتری تولید شود، بایستی از رابطه بالا استفاده کنیم.

ب

\(\begin{array}{l}m\left( t \right) = 5000 \Rightarrow t = {\log _2}5000 = {\log _2}{5^4} \times {2^3}\\\\ = 4{\log _2}5 + 3{\log _2}2 = 4{\log _2}\frac{{10}}{2} + 3\end{array}\)

\( = 4\left( {{{\log }_2}10 - {{\log }_2}2} \right) + 3\)  

ساعت  \( = 4{\log _2}10 - 1 = 4 \times \frac{1}{{{{\log }_{10}}2}} - 1 \simeq \frac{4}{{0/301}} - 1 \simeq 12/5 \equiv 12:30'\)

الف

\(\left( {b \ne 1\;,\;a\;,\;b > 0} \right)\;\;\;{a^{{{\log }_b}a}} = 0\) ×

\(\begin{array}{l}{a^{{{\log }_b}a}} = a\mathop \Rightarrow \limits^{\left( {{{\log }_b}} \right)} {\log _b}\left( {{a^{{{\log }_b}a}}} \right) = {\log _b}a \Rightarrow {\log _b}a \times {\log _b}a = {\log _b}a\\\\ \Rightarrow {\log _b}a = 1 \Rightarrow a = b \times \end{array}\)

ب 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\left( {d \ne 1\;,\;a\;,\;b\;,\;c\;,\;d > 0} \right)\;\;\;{\log _d}abc = {\log _d}a + {\log _d}b + {\log _d}c\)  ✔

\(\begin{array}{l}{\log _d}abc = {\log _d}a\left( {bc} \right) = {\log _d}a + {\log _d}bc\\\\ = {\log _d}a + {\log _d}b + {\log _d}c\end{array}\)

پ

\(\log x\;\log y = \log x + \log y\) ×

\(\begin{array}{l}\log x\;\log y = \log x + \log y\\\\ \Rightarrow \log {x^{\log y}} = \log xy \Rightarrow xy = {x^{\log y}} \times \end{array}\)

ت

لگاریتم هر عدد مثبت، همواره عددی مثبت است. ×

\(\log \left( {0/1} \right) = \log {10^{ - 1}} = - \log 10 = - 1\)

الف

\(m\left( t \right) = 1 \times {2^{ - \;\frac{t}{4}}} = {2^{ - \;\frac{t}{4}}}\)

ب

\(\begin{array}{l}\left\{ \begin{array}{l}m\left( t \right) = {2^{ - \;\frac{t}{4}}}\\m\left( t \right) = 0/01\end{array} \right. \Rightarrow 0/01 = {2^{ - \;\frac{t}{4}}}\\\\ \Rightarrow \log {10^{ - 2}} = \log {2^{ - \;\frac{t}{4}}} = - \;\frac{t}{4}\log 2\\\\ \Rightarrow - 2 = - \;\frac{t}{4}\log 2\end{array}\)

روز  \( \Rightarrow t = \frac{8}{{\log 2}} \simeq \frac{8}{{0/301}} \simeq 26/5\)

الف

\(\begin{array}{l}\log \left( {18 \times 375} \right) = \log \left( {2 \times {3^2} \times 3 \times {5^3}} \right)\\\\ = \log \left( {2 \times {3^3} \times {{\left( {\frac{{10}}{2}} \right)}^3}} \right) = \log \left( {{2^{ - 2}} \times {3^3} \times {{10}^3}} \right)\\\\ = - 2\log 2 + 3\log 3 + 3\log 10\\\\ \simeq - 2 \times 0/301 + 3 \times 0/4771 + 3 = 3/8293\end{array}\)

ب

\(\begin{array}{l}\log \sqrt {0/75} = \frac{1}{2}\log \frac{3}{4} = \frac{1}{2}\left( {\log 3 - \log {2^2}} \right) = \frac{1}{2}\left( {\log 3 - 2\log 2} \right)\\\\ = \frac{1}{2}\left( {0/4771 - 2 \times 0/301} \right)\\\\ = - 0/06245\end{array}\)

پ

\(\begin{array}{l}{\log _2}\frac{{\sqrt 8 }}{{\sqrt[4]{2}}} = {\log _2}\sqrt {{2^3}} - {\log _2}\sqrt[4]{2} = {\log _2}{2^{\frac{3}{2}}} - {\log _2}{2^{\frac{1}{4}}}\\\\ = \frac{3}{2}{\log _2}2 - \frac{1}{4}{\log _2}2\\\\ = \frac{3}{2} - \frac{1}{4} = \frac{5}{4}\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}f(x) = {\log _a}x\\(\frac{1}{2}\;,\;4)\end{array} \right\} \Rightarrow f(\frac{1}{2}) = 4 \Rightarrow 4 = {\log _a}\frac{1}{2}\\\\ \Rightarrow {a^4} = \frac{1}{2}\,\,\,\,\,\,\mathop \Rightarrow \limits^{a > 0\atop a \ne 1} \,\,\,\,\,a = \sqrt[4]{{\frac{1}{2}}} = \frac{1}{{\sqrt[4]{2}}} = \frac{{\sqrt[4]{8}}}{2}\end{array}\)

\({\log _b}a \times {\log _a}b = 1\) ✔

\(\begin{array}{l}{b^{{{\log }_b}a \times {{\log }_a}b}} = {b^1} = b \Rightarrow \\{({b^{{{\log }_b}a}})^{{{\log }_a}b}} = {a^{{{\log }_a}b}} = b\end{array}\)

\(\log \;5 = \log \;3 + \log \;2\) ×

\(\begin{array}{l}\log 5 = \log (3 \times 2) = \log 6\\ \Rightarrow 5 = 6\mathop {}\nolimits^{} \times \end{array}\)

\(\begin{array}{l}m\left( t \right) = 128 \times {2^{ - \;\frac{t}{{30}}}} \Rightarrow m\left( {300} \right) = 128 \times {2^{ - \;\frac{{300}}{{30}}}}\\\\ = {2^7} \times {2^{ - 10}} = {2^{ - \,3}} = 0/125\end{array}\)