1.9-bo'lim : Eksponensial va logarifm tenglamalari
Eksponensial o'sish/emirilish. Dunyodagi ko'plab miqdorlarni eksponensial o'sish/emirilish tenglamasi bilan modellashtirish mumkin (hech bo'lmaganda qisqa vaqt).
\[Q = {Q_0}{{\bf{e}}^{k\,t}}\]Agar \(k\) ijobiy bo'lsa, biz eksponensial o'sishni olamiz va \(k\) manfiy bo'lsa, biz eksponensial parchalanishni olamiz.
16. Bizda dastlab 100 gramm radioaktiv element bor va 1250 yildan keyin 80 gramm qoladi.
- Ushbu element uchun eksponensial yemirilish tenglamasini aniqlang.
- Elementning yarmi parchalanishi uchun qancha vaqt kerak bo'ladi?
- Elementdan atigi 1 gramm qolguncha qancha vaqt ketadi?
a Ushbu element uchun eksponensial yemirilish tenglamasini aniqlang. Yechim
Biz buni bilishimizni tan olishdan boshlashimiz mumkin,
\[Q\left( 0 \right) = 100\hspace{0.5in}{\rm{and}}\hspace{0.5in}Q\left( {1250} \right) = 80\]Agar biz tenglamada birinchi shartdan foydalansak,
\[100 = Q\left( 0 \right) = {Q_0}{{\bf{e}}^{k\left( 0 \right)}} = {Q_0}\hspace{0.5in} \to \hspace{0.5in}{Q_0} = 100\]Endi biz tenglamadagi birinchi noma'lumni bilamiz. Buni va ikkinchi shartni tenglamaga qo'shish bizga beradi,
\[80 = Q\left( {1250} \right) = 100{{\bf{e}}^{1250k}}\]\(k\) qiymatini aniqlash uchun biz ushbu bo'limdagi oldingi masalalardagi usullardan foydalanishimiz mumkin.
\[\begin{align*}80 & = 100{{\bf{e}}^{1250k}}\\ \frac{{80}}{{100}} & = {{\bf{e}}^{1250k}}\\ \ln \left( {\frac{4}{5}} \right) & = 1250k\\ k & = \frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right) = - 0.000178515\end{align*}\]Sizning afzalliklaringizga qarab, biz aniq qiymat yoki kasr qiymatidan foydalanishimiz mumkin. Shuni yodda tutingki, \(k\) eksponensial funktsiyaning ko'rsatkichida bo'lganligi sababli, agar biz haddan tashqari yaxlitlashimiz mumkin bo'lgan qiymatdagi potentsial katta farqlarga yo'l qo'ymaslik uchun bir nechta o'nli kasrlardan foydalanishimiz kerak.
Bularning barchasini jamlagan holda, ushbu populyatsiya uchun eksponensial yemirilish tenglamasi,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{Q = 100{{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)t}}}}\]b Elementning yarmi parchalanishi uchun qancha vaqt kerak bo'ladi? Yechim
Bu erda bizdan haqiqatan ham so'ralayotgan narsa tenglamani yechishdir,
\[50 = Q\left( t \right) = 100{{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)t}}\]va biz buni qanday qilishni ushbu bo'limdagi oldingi muammolardan bilamiz. Mana bu qism uchun yechim ishi.
\[\begin{align*}\frac{{50}}{{100}} & = {{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)t}}\\ \ln \left( {\frac{1}{2}} \right) & = \frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)t\\ t & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{1250\ln \left( {{\textstyle{1 \over 2}}} \right)}}{{\ln \left( {{\textstyle{4 \over 5}}} \right)}} = 3882.8546}}\end{align*}\]Elementning yarmi parchalanishi uchun 3882,8546 yil kerak bo'ladi. Yon eslatmada bu vaqt deyiladi yarim yemirilish davri elementdan.
c Elementdan atigi 1 gramm qolguncha qancha vaqt ketadi? Yechim
Ushbu qismda bizdan tenglamani yechish so'raladi,
\[1 = Q\left( t \right) = 100{{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)t}}\]va biz buni qanday qilishni ushbu bo'limdagi oldingi muammolardan bilamiz. Mana bu qism uchun yechim ishi.
\[\begin{align*}\frac{1}{{100}} & = {{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)t}}\\ \ln \left( {\frac{1}{{100}}} \right) & = \frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)t\\ t & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{1250\ln \left( {{\textstyle{1 \over {100}}}} \right)}}{{\ln \left( {{\textstyle{4 \over 5}}} \right)}} = 25797.1279}}\end{align*}\]25 797,1279 yildan keyin faqat 1 gramm element qoladi.