Thu. Jul 18th, 2024

Apr 25, 2022

# Koshe dekhi 21class 9

### Koshe dekhi 21 class 9

 1. মান নির্ণয় করি : (i) ${\color{Blue}&space;log_{4}\left&space;(&space;\frac{1}{64}&space;\right&space;)}$ সমাধানঃ $log_{4}\left&space;(&space;\frac{1}{64}&space;\right&space;)$ $=-log_{4}\,&space;64\;&space;{\color{Blue}&space;[\because&space;log\left&space;(&space;\frac{1}{a}&space;\right&space;)=-log\,&space;a]}$ $=-log_{4}\,&space;4^{3}$ $=-3\,&space;log_{4}\,&space;4\;&space;{\color{Blue}&space;[\because&space;log\,&space;a^{m}=mlog\,&space;a]}$ = −3 × (1) ${\color{Blue}&space;[\because&space;log_{a}\,&space;a=1]}$ = −3 (উত্তর)

 1. মান নির্ণয় করি : (ii) ${\color{Blue}&space;log_{0.01}0.000001}$ সমাধানঃ $log_{0.01}0.000001$ $=\frac{log\,&space;0.000001}{log\,&space;0.01}$ ${\color{Blue}&space;[\because&space;log_{a}\,&space;b=\frac{logb}{loga}]}$ $=\frac{log\,&space;\left&space;(&space;\frac{1}{1000000}&space;\right&space;)}{log\,&space;\left&space;(\frac{1}{100}&space;\right&space;)}$ $=\frac{-log\,&space;100000}{-log\,&space;100}\;&space;{\color{Blue}&space;[\because&space;log\,&space;\left&space;(&space;\frac{1}{a}&space;\right&space;)=-log\,&space;a]}$ $=\frac{log\,&space;10^{6}}{log\,&space;10^{2}}$ $=\frac{6\,&space;log\,&space;10}{2\,&space;log\,&space;10}\;&space;{\color{Blue}&space;[\because&space;log\,&space;a^{m}=m\,&space;log\,&space;a]}$ = 3 (উত্তর) অথবা  $log_{0.01}0.000001$ $=log_{0.01}\left&space;(&space;0.01&space;\right&space;)^{3}$ = 3 log0.01 0.01 ${\color{Blue}&space;[\because&space;log\,&space;a^{m}=m\,&space;log\,&space;a]}$ = 3 ${\color{Blue}&space;[\because&space;log_{a}\,&space;a=1]}$ (উত্তর)

 1. মান নির্ণয় করি : (iii) ${\color{Blue}&space;log_{\sqrt{6}}216}$ সমাধানঃ $log_{\sqrt{6}}216$ $=\frac{log\,&space;216}{log\,&space;\sqrt{6}}$ ${\color{Blue}&space;[\because&space;log_{a}\,&space;b=\frac{logb}{loga}]}$ $=\frac{log\,&space;6^{3}}{log\,&space;6^{\frac{1}{2}}}$ $=\frac{3\,&space;log\,&space;6}{\frac{1}{2}\,&space;log\,&space;6}$ ${\color{Blue}&space;[\because&space;log\,&space;a^{m}=m\,&space;log\,&space;a]}$ $=\frac{3}{\left&space;(\frac{1}{2}&space;\right&space;)}$ $=3\times&space;\frac{2}{1}$ = 6 (উত্তর)

 1. মান নির্ণয় করি : (iv) ${\color{Blue}&space;log_{2\sqrt{3}}1728}$ সমাধানঃ $log_{2\sqrt{3}}1728$ $=\frac{log\,&space;1728}{log\,&space;2\sqrt{3}}$ ${\color{Blue}&space;[\because&space;log_{a}\,&space;b=\frac{logb}{loga}]}$ ∴ 1728 = 26 × 33 = 26 × (√3)6 = (2√3)6 এখন, $\frac{log\,&space;1728}{log\,&space;2\sqrt{3}}$ $=\frac{log\,&space;\left&space;(&space;2\sqrt{3}&space;\right&space;)^{6}}{log\,&space;\left&space;(&space;2\sqrt{3}&space;\right&space;)}$ $=\frac{6\,&space;log\,&space;2\sqrt{3}}{log\,&space;2\sqrt{3}}$ = 6 (উত্তর)

 2. (a) 625 -এর লগারিদম 4 হলে, নিধান কি হবে হিসাব করে লিখি।   সমাধানঃ  ধরি, নিধান হল  ‘x‘ ∴ logx 625 = 4 বা, x4 = 625 [যেহেতু, loga b = c হলে, ac = b হয়] বা, x4 = 5 × 5 × 5 × 5 বা, x4 = 54 ∴ x = 5  উত্তরঃ নির্ণেয় নিধান হল 5

 2. (b) 5832 -এর লগারিদম 6 হলে, নিধান কি হবে হিসাব করে লিখি।   সমাধানঃ ধরি, নিধান হলো  ‘x‘ ∴ logx 5832 = 6 বা, x6 = 5832 [যেহেতু, loga b = c হলে, ac = b হয়] বা, x6 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 বা, x6 = (3√2)6 ∴ x = 3√2 উত্তরঃ নির্ণেয় নিধান হল 3√2

 3. (a)  ${\color{Blue}&space;1+log_{10}\,&space;a=2\,&space;log_{10}\,&space;b}$ হলে, a কে b -এর দ্বারা প্রকাশ করি।  সমাধানঃ $1+log_{10}\,&space;a=2\,&space;log_{10}\,&space;b$ বা, 1 = 2 log10 b − log10 a বা, log10 10 = log10 b2 − log10 a  ${\color{Blue}&space;[\because&space;log_{a}\,&space;a=1]}$ এবং ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ বা, $log\,&space;_{10}\,&space;10=log\,&space;_{10}\,\frac{b^{2}}{a}$ ${\color{Blue}&space;\left&space;[&space;\because&space;log\,&space;_{a}\,&space;m-log\,&space;_{a}\,&space;n=log\,&space;_{a}\,&space;\frac{m}{n}&space;\right&space;]}$ বা, $10=\frac{b^{2}}{a}$ ${\color{DarkGreen}&space;\therefore&space;a=\frac{b^{2}}{10}}$ (উত্তর)

 3. (b) ${\color{Blue}&space;3+log\,&space;_{10}\,&space;x=2\,&space;log\,&space;_{10}\,&space;y}$ হলে, x কে y -এর দ্বারা প্রকাশ করি।  সমাধানঃ $3+log\,&space;_{10}\,&space;x=2\,&space;log\,&space;_{10}\,&space;y$ বা, $log\,&space;_{10}\,&space;x=2\,&space;log\,&space;_{10}\,&space;y-3$ বা, $log\,&space;_{10}\,&space;x=2\,&space;log\,&space;_{10}\,&space;y-3$ × 1 বা, $log\,&space;_{10}\,&space;x=2\,&space;log\,&space;_{10}\,&space;y-3\,&space;log\,&space;_{10}\,&space;10$  ${\color{Blue}&space;[\because&space;log_{a}\,&space;a=1]}$ বা, $log\,&space;_{10}\,&space;x=log\,&space;_{10}\,&space;y^{2}-log\,&space;_{10}\,&space;10^{3}$  ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ বা, $log\,&space;_{10}\,&space;x=log\,&space;_{10}\,\frac{y^{2}}{10^{3}}$ ${\color{Blue}&space;\left&space;[&space;\because&space;log\,&space;_{a}\,&space;m-log\,&space;_{a}\,&space;n=log\,&space;_{a}\,&space;\frac{m}{n}&space;\right&space;]}$ ${\color{DarkGreen}&space;\therefore&space;x=\frac{y^{2}}{1000}}$   (উত্তর)

 4. মান নির্ণয় করি : (a) ${\color{Blue}&space;log_{2}\left&space;[&space;log_{2}\left&space;\{&space;log_{3}\left&space;(&space;log_{3}27^{3}&space;\right&space;)&space;\right&space;\}&space;\right&space;]}$ সমাধানঃ $log_{2}\left&space;[&space;log_{2}\left&space;\{&space;log_{3}\left&space;(&space;log_{3}\,&space;27^{3}&space;\right&space;)&space;\right&space;\}&space;\right&space;]$ $=log_{2}\left&space;[&space;log_{2}\left&space;\{&space;log_{3}\left&space;(&space;log_{3}\,&space;3^{3^{3}}&space;\right&space;)&space;\right&space;\}&space;\right&space;]$ $=log_{2}\left&space;[&space;log_{2}\left&space;\{&space;log_{3}\left&space;(&space;log_{3}\,&space;3^{9}&space;\right&space;)&space;\right&space;\}&space;\right&space;]$ $=log_{2}\left&space;[&space;log_{2}\left&space;\{&space;log_{3}\left&space;(&space;9\,&space;log_{3}\,&space;3&space;\right&space;)&space;\right&space;\}&space;\right&space;]$ ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ $=log_{2}\left&space;[&space;log_{2}\left&space;\{&space;log_{3}\left&space;(&space;9\times&space;1&space;\right&space;)&space;\right&space;\}&space;\right&space;]$ ${\color{Blue}&space;[\because&space;log_{a}\,&space;a=1]}$ $=log_{2}\left&space;[&space;log_{2}\left&space;\{&space;log_{3}\,&space;9&space;\right&space;\}&space;\right&space;]$ $=log_{2}\left&space;[&space;log_{2}\left&space;\{&space;log_{3}\,&space;3^{2}&space;\right&space;\}&space;\right&space;]$ $=log_{2}\left&space;[&space;log_{2}\left&space;\{&space;2\,&space;log_{3}\,&space;3&space;\right&space;\}&space;\right&space;]$ ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ $=log_{2}\left&space;[&space;log_{2}\left&space;\{&space;2\times&space;1\right&space;\}&space;\right&space;]$ ${\color{Blue}&space;[\because&space;log_{a}\,&space;a=1]}$ $=log_{2}\left&space;[&space;log_{2}\,&space;2&space;\right&space;]$ $=log_{2}\,&space;1$ = 0 ${\color{Blue}&space;\left&space;[\because&space;log_{x}\,&space;1=0&space;\right&space;]}$ (উত্তর)

 4. মান নির্ণয় করি : (b) ${\color{Blue}&space;\frac{log\sqrt{27}+log8-log\sqrt{1000}}{log1.2}}$ সমাধানঃ $\frac{log\sqrt{27}+log8-log\sqrt{1000}}{log1.2}$ $=\frac{log\,&space;\sqrt{3^{3}}+log\,&space;2^{3}-log\,&space;\sqrt{10^{3}}}{log\,&space;\frac{12}{10}}$ $=\frac{log\,&space;3^{\frac{3}{2}}+3\,&space;log\,&space;2-log\,&space;10^{\frac{3}{2}}}{log\,&space;\frac{3\times&space;4}{10}}$ ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ $=\frac{\frac{3}{2}\,&space;log\,&space;3+3\,&space;log\,&space;2-\frac{3}{2}\,&space;log\,&space;10}{log\,&space;3+log\,&space;4-log\,&space;10}$ ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$,  ${\color{Blue}&space;\left&space;[&space;\because&space;log\,&space;_{a}\,&space;mn=log\,&space;_{a}\,&space;m+log\,&space;_{a}\,&space;n&space;\right&space;]}$ এবং ${\color{Blue}&space;\left&space;[&space;\because&space;log\,&space;_{a}\,&space;\frac{m}{n}=log\,&space;_{a}\,&space;m-log\,&space;_{a}\,&space;n&space;\right&space;]}$ $=\frac{\frac{3}{2}\,&space;log\,&space;3+\frac{3}{2}\times&space;2\,&space;log\,&space;2-\frac{3}{2}\,&space;log\,&space;10}{log\,&space;3+log\,&space;2^{2}-log\,&space;10}$ $=\frac{\frac{3}{2}\left&space;(log\,&space;3+2\,&space;log\,&space;2-log\,&space;10&space;\right&space;)}{\left&space;(log\,&space;3+2\,&space;log\,&space;2-log\,&space;10&space;\right&space;)}$ ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ ${\color{DarkGreen}&space;=\frac{3}{2}}$ (উত্তর)

 4. মান নির্ণয় করি : (c) ${\color{Blue}&space;log_{3}\,&space;4\times&space;log_{4}\,&space;5\times&space;log_{5}\,&space;6\times&space;log_{6}\,&space;7\times&space;log_{7}\,&space;3}$ সমাধানঃ $log_{3}\,&space;4\times&space;log_{4}\,&space;5\times&space;log_{5}\,&space;6\times&space;log_{6}\,&space;7\times&space;log_{7}\,&space;3$ $=\frac{log\,&space;4}{log\,&space;3}\times&space;\frac{log\,&space;5}{log\,&space;4}\times&space;\frac{log\,&space;6}{log\,&space;5}\times&space;\frac{log\,&space;7}{log\,&space;6}\times&space;\frac{log\,&space;3}{log\,&space;7}$  [ধরি, প্রতিটির log এর নিধান একই] = 1 (উত্তর)

 4. মান নির্ণয় করি : (d) ${\color{Blue}&space;log_{10}\frac{384}{5}+log_{10}\frac{81}{32}+3log_{10}\frac{5}{3}+log_{10}\frac{1}{9}}$ সমাধানঃ $log_{10}\,&space;\frac{384}{5}+log_{10}\,&space;\frac{81}{32}+3log_{10}\,&space;\frac{5}{3}+log_{10}\,&space;\frac{1}{9}$ $=log_{10}\,&space;384-log_{10}\,&space;5+log_{10}\,&space;81-log_{10}\,&space;32+3log_{10}\,&space;5-3log_{10}\,&space;3+log_{10}\,&space;1-log_{10}\,&space;9$ $=log_{10}\,&space;2^{7}\times&space;3-log_{10}\,&space;5+log_{10}\,&space;3^{4}-log_{10}\,&space;2^{5}+3log_{10}\,&space;5-3log_{10}\,&space;3+log_{10}\,&space;1-log_{10}\,&space;3^{2}$ $=7log_{10}\,&space;2+log_{10}\,&space;3-log_{10}\,&space;5+4log_{10}\,&space;3-5log_{10}\,&space;2+3log_{10}\,&space;5-3log_{10}\,&space;3+log_{10}\,&space;1-2log_{10}\,&space;3$ $=2log_{10}\,&space;2+2log_{10}\,&space;5$ ${\color{Blue}&space;\left&space;[\because&space;log_{x}\,&space;1=0&space;\right&space;]}$ $=log_{10}\,&space;2^{2}+log_{10}\,&space;5^{2}$ ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ $=log_{10}\,&space;4+log_{10}\,&space;25$ $=log_{10}\,&space;\left&space;(4\times&space;25&space;\right&space;)$ ${\color{Blue}&space;\left&space;[&space;\because&space;log\,&space;_{a}\,&space;mn=log\,&space;_{a}\,&space;m+log\,&space;_{a}\,&space;n&space;\right&space;]}$ $=log_{10}\,&space;100$ $=log_{10}\,&space;10^{2}$ $=2log_{10}\,&space;10$ ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ = 2  (উত্তর)

 5. প্রমাণ করি : (i) ${\color{Blue}&space;log\frac{75}{16}-2log\frac{5}{9}+log\frac{32}{243}=log2}$ সমাধানঃ  বামপক্ষ : $log\frac{75}{16}-2log\frac{5}{9}+log\frac{32}{243}$ $=log\,&space;75-log\,&space;16-2log\,&space;5+2log\,&space;9+log\,&space;32-log\,&space;243$ $=log\,&space;\left&space;(5^{2}\times&space;3&space;\right&space;)-log\,&space;2^{4}-2log\,&space;5+2log\,&space;3^{2}+log\,&space;2^{5}-log\,&space;3^{5}$ $=log\,&space;5^{2}+log\,&space;3-log\,&space;2^{4}-2log\,&space;5+2\times&space;2log\,&space;3+log\,&space;2^{5}-log\,&space;3^{5}$ ${\color{Blue}&space;\left&space;[&space;\because&space;log\,&space;_{a}\,&space;mn=log\,&space;_{a}\,&space;m+log\,&space;_{a}\,&space;n&space;\right&space;]}$ $=2log\,&space;5+log\,&space;3-4log\,&space;2-2log\,&space;5+4log\,&space;3+5log\,&space;2-5log\,&space;3$ ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ = log 2 = ডানপক্ষ (প্রমাণিত)

 5. প্রমাণ করি : (ii) ${\color{Blue}&space;log_{10}15\left&space;(&space;1+log_{15}30&space;\right&space;)+\frac{1}{2}log_{10}16\left&space;(&space;1+log_{4}7\right&space;)-log_{10}6\left&space;(&space;log_{6}3+1+log_{6}7\right&space;)=2}$ সমাধানঃ বামপক্ষ : $log_{10}15\left&space;(&space;1+log_{15}30&space;\right&space;)+\frac{1}{2}log_{10}16\left&space;(&space;1+log_{4}7\right&space;)-log_{10}6\left&space;(&space;log_{6}3+1+log_{6}7\right&space;)$   $=log_{10}\,&space;15+log_{10}\,&space;15\times&space;log_{15}\,&space;30&space;+\frac{1}{2}log_{10}\,&space;16+\frac{1}{2}log_{10}\,&space;16\times&space;log_{4}\,&space;7-log_{10}\,&space;6\times&space;log_{6}\,&space;3-log_{10}\,&space;6-log_{10}\,&space;6\times&space;log_{6}\,&space;7$   $=log_{10}\,&space;15+\left&space;(\frac{log\,&space;15}{log\,&space;10}\times&space;\frac{log\,&space;30}{log\,&space;15}&space;\right&space;)+\frac{1}{2}log_{10}\,&space;4^{2}+\frac{1}{2}log_{10}\,&space;4^{2}\times&space;log_{4}\,&space;7-\left&space;(\frac{log\,&space;6}{log\,&space;10}\times&space;\frac{log\,&space;3}{log\,&space;6}&space;\right&space;)-log_{10}\,&space;6-\left&space;(\frac{log\,&space;6}{log\,&space;10}\times&space;\frac{log\,&space;7}{log\,&space;6}&space;\right&space;)$   $=log_{10}\,&space;15+log_{10}\,&space;30&space;+\frac{1}{2}\times&space;2\,&space;log_{10}\,&space;4+\frac{1}{2}\times&space;2\,&space;log_{10}\,&space;4\times&space;log_{4}\,&space;7-log_{10}\,&space;3-log_{10}\,&space;6-log_{10}\,&space;7$   $=log_{10}\,&space;15+log_{10}\,&space;30&space;+log_{10}\,&space;4+\left&space;(&space;\frac{log\,&space;4}{log\,&space;10}\times&space;\frac{log\,&space;7}{log\,&space;4}&space;\right&space;)-log_{10}\,&space;3-log_{10}\,&space;6-log_{10}\,&space;7$   $=log_{10}\,&space;15+log_{10}\,&space;30&space;+log_{10}\,&space;4+log_{10}\,&space;7-log_{10}\,&space;3-log_{10}\,&space;6-log_{10}\,&space;7$   $=\left&space;(log_{10}\,&space;15+log_{10}\,&space;30&space;+log_{10}\,&space;4+log_{10}\,&space;7&space;\right&space;)-\left&space;(log_{10}\,&space;3+log_{10}\,&space;6+log_{10}\,&space;7&space;\right&space;)$   $=log_{10}\,&space;\frac{15\times&space;30\times&space;4\times&space;7}{3\times&space;6\times&space;7}$ $=log_{10}\,&space;100$ $=log_{10}\,&space;10^{2}$ $=2log_{10}\,&space;10$ = 2 = ডানপক্ষ (প্রমাণিত)

 5. প্রমাণ করি : (iii) ${\color{Blue}&space;log_{2}log_{2}log_{4}\,&space;256+2log_{\sqrt{2}}\,&space;2=5}$ সমাধানঃ বামপক্ষ : $log_{2}log_{2}log_{4}\,&space;256+2log_{\sqrt{2}}\,&space;2$ $=log_{2}log_{2}log_{4}\,&space;4^{4}+2log_{\sqrt{2}}\,&space;\left&space;(\sqrt{2}&space;\right&space;)^{2}$ $=log_{2}\,&space;log_{2}\,&space;4log_{4}\,&space;4+2\times&space;2\,&space;log_{\sqrt{2}}\,&space;\sqrt{2}$  ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ $=log_{2}\,&space;log_{2}\,&space;4\times&space;1+4\times&space;1$ ${\color{Blue}&space;[\because&space;log_{a}\,&space;a=1]}$ $=log_{2}\,&space;log_{2}\,&space;4+4$ $=log_{2}\,&space;log_{2}\,&space;2^{2}+4$ $=log_{2}\,&space;2log_{2}\,&space;2+4$ ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ $=log_{2}\,&space;2\times&space;1+4$ ${\color{Blue}&space;[\because&space;log_{a}\,&space;a=1]}$ $=log_{2}\,&space;2+4$ = 1 + 4  ${\color{Blue}&space;[\because&space;log_{a}\,&space;a=1]}$ = 5 = ডানপক্ষ (প্রমাণিত)

 5. প্রমাণ করি : (iv) ${\color{Blue}&space;log_{x^{2}}\,&space;x\times&space;log_{y^{2}}\,&space;y\times&space;log_{z^{2}}\,&space;z=\frac{1}{8}}$ সমাধানঃ বামপক্ষ : $log_{x^{2}}\,&space;x\times&space;log_{y^{2}}\,&space;y\times&space;log_{z^{2}}\,&space;z$ $=log_{x^{2}}\,&space;\left&space;(x^{2}&space;\right&space;)^{\frac{1}{2}}\times&space;log_{y^{2}}\,&space;\left&space;(y^{2}&space;\right&space;)^{\frac{1}{2}}\times&space;log_{z^{2}}\,&space;\left&space;(z^{2}&space;\right&space;)^{\frac{1}{2}}$ $=\frac{1}{2}\,&space;log_{x^{2}}\,&space;x^{2}\times&space;\frac{1}{2}\,&space;log_{y^{2}}\,&space;y^{2}\times&space;\frac{1}{2}\,&space;log_{z^{2}}\,&space;z^{2}$ ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ $=\frac{1}{2}\times&space;1\times&space;\frac{1}{2}\times&space;1\times&space;\frac{1}{2}\times&space;1$  ${\color{Blue}&space;[\because&space;log_{a}\,&space;a=1]}$ $=\frac{1}{8}$ = ডানপক্ষ (প্রমাণিত)

 5. প্রমাণ করি : (v) ${\color{Blue}&space;log_{b^{3}}\,&space;a\times&space;log_{c^{3}}\,&space;b\times&space;log_{a^{3}}\,&space;c=\frac{1}{27}}$ সমাধানঃ বামপক্ষ : $log_{b^{3}}\,&space;a\times&space;log_{c^{3}}\,&space;b\times&space;log_{a^{3}}\,&space;c$ $=\frac{log\,&space;a}{log\,&space;b^{3}}\times&space;\frac{log\,&space;b}{log\,&space;c^{3}}\times&space;\frac{log\,&space;c}{log\,&space;a^{3}}$ $=\frac{log\,&space;a}{3\,&space;log\,&space;b}\times&space;\frac{log\,&space;b}{3\,&space;log\,&space;c}\times&space;\frac{log\,&space;c}{3\,&space;log\,&space;a}$ $=\frac{1}{3}\times&space;\frac{1}{3}\times&space;\frac{1}{3}$ $=\frac{1}{27}$ = ডানপক্ষ (প্রমাণিত)

Koshe dekhi 21 class 9

 5. প্রমাণ করি : (vi) ${\color{Blue}&space;\frac{1}{log_{xy}\left&space;(&space;xyz&space;\right&space;)}+\frac{1}{log_{yz}\left&space;(&space;xyz&space;\right&space;)}+\frac{1}{log_{zx}\left&space;(&space;xyz&space;\right&space;)}=2}$ সমাধানঃ বামপক্ষ : $\frac{1}{log_{xy}\left&space;(&space;xyz&space;\right&space;)}+\frac{1}{log_{yz}\left&space;(&space;xyz&space;\right&space;)}+\frac{1}{log_{zx}\left&space;(&space;xyz&space;\right&space;)}$ $=\frac{log\,&space;xy}{log\,&space;xyz}+\frac{log\,&space;yz}{log\,&space;xyz}+\frac{log\,&space;zx}{log\,&space;xyz}$ $=\frac{log\,&space;xy+log\,&space;yz+log\,&space;zx}{log\,&space;xyz}$ $=\frac{log\,&space;\left&space;(xy\times&space;yz\times&space;zx&space;\right&space;)}{log\,&space;xyz}$  ${\color{Blue}&space;\left&space;[&space;\because&space;log\,&space;_{a}\,&space;mn=log\,&space;_{a}\,&space;m+log\,&space;_{a}\,&space;n&space;\right&space;]}$ $=\frac{log\,&space;\left&space;(x^{2}y^{2}z^{2}&space;\right&space;)}{log\,&space;xyz}$ $=\frac{log\,&space;\left&space;(xyz&space;\right&space;)^{2}}{log\,&space;xyz}$ $=\frac{2\,&space;log\,&space;xyz}{log\,&space;xyz}$  ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ = 2 = ডানপক্ষ (প্রমাণিত)

Koshe dekhi 21 class 9

 5. প্রমাণ করি : (vii) ${\color{Blue}&space;log\frac{a^{2}}{bc}+log\frac{b^{2}}{ca}+log\frac{c^{2}}{ab}=0}$ সমাধানঃ বামপক্ষ : $log\frac{a^{2}}{bc}+log\frac{b^{2}}{ca}+log\frac{c^{2}}{ab}$ $=log\,&space;a^{2}-log\,&space;bc+log\,&space;b^{2}-log\,&space;ca+log\,&space;c^{2}-log\,&space;ab$ $=\left&space;(log\,&space;a^{2}+log\,&space;b^{2}+log\,&space;c^{2}&space;\right&space;)-\left&space;(log\,&space;bc+log\,&space;ca+log\,&space;ab&space;\right&space;)$ $=log\,&space;\left&space;(a^{2}\times&space;b^{2}\times&space;c^{2}&space;\right&space;)-log\,&space;\left&space;(&space;bc\times&space;ca\times&space;ab&space;\right&space;)$ $=log\left&space;(&space;a^{2}b^{2}c^{2}&space;\right&space;)-log\left&space;(&space;a^{2}b^{2}c^{2}&space;\right&space;)$ = 0 = ডানপক্ষ (প্রমাণিত)

Koshe dekhi 21 class 9

 5. প্রমাণ করি : (viii) xlog y − log z × ylog z − log x × zlog x − log y = 1 সমাধানঃ ধরি, k = xlog y − log z × ylog z − log x × zlog x − log y  উভয়পক্ষে log নিয়ে পাই, log k = log (xlog y − log z × ylog z − log x × zlog x − log y) বা, log k = (log y − log z) log x + (log z − log x) log y + (log x − log y) log z বা, log k = log y log x − log z log x + log z log y − log y log x + log z log x − log z log y বা, log k = 0 বা, log k = log 1 ∴ k = 1 ∴ xlog y − log z × ylog z − log x × zlog x − log y = 1 বামপক্ষ = ডানপক্ষ (প্রমাণিত)

Koshe dekhi 21 class 9

 6.(i) যদি  ${\color{Blue}&space;log\,&space;\frac{x+y}{5}=\frac{1}{2}\left&space;(&space;log\,&space;x+log\,&space;y&space;\right&space;)}$   হয়, তাহলে দেখাই যে  ${\color{Blue}&space;\frac{x}{y}+\frac{y}{x}=23}$ সমাধানঃ  $log\,&space;\frac{x+y}{5}=\frac{1}{2}\left&space;(&space;log\,&space;x+log\,&space;y&space;\right&space;)$ বা, $2\,&space;log\,&space;\frac{x+y}{5}=\left&space;(&space;log\,&space;x+log\,&space;y&space;\right&space;)$ বা, $log\,&space;\left&space;(\frac{x+y}{5}&space;\right&space;)^{2}=log\,&space;xy$ বা, $\left&space;(\frac{x+y}{5}&space;\right&space;)^{2}=xy$ বা, $\frac{x^{2}+2xy+y^{2}}{25}=xy$ বা, x2 + 2xy + y2 = 25xy বা, x2 + y2 = 25xy − 2xy  বা, x2 + y2 = 23xy বা, $\frac{x^{2}+y^{2}}{xy}=23$ বা, $\frac{x^{2}}{xy}+\frac{y^{2}}{xy}=23$ ${\color{DarkGreen}&space;\therefore&space;\frac{x}{y}+\frac{y}{x}=23}$ (প্রমাণিত)

Koshe dekhi 21 class 9

 6(ii) যদি  a4 + b4 = 14a2 b2  হয়, তাহলে দেখাই যে log (a2 + b2 ) = log a + log b + 2 log 2 সমাধানঃ a4 + b4 = 14a2 b2  বা, (a2 + b2 )2 − 2a2 b2 = 14a2 b2  বা, (a2 + b2 )2 = 14a2 b2 + 2a2 b2  বা, (a2 + b2 )2 = 16a2 b2 বা, (a2 + b2 )2 = (4ab)2 বা, a2 + b2 = 4ab উভয়পক্ষে log নিয়ে পাই, log (a2 + b2) = log (4ab) বা, log (a2 + b2) = log 4 + log a + log b বা, log (a2 + b2) = log 22 + log a + log b বা, log (a2 + b2) = 2 log 2 + log a + log b  ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ ∴ log (a2 + b2 ) = log a + log b + 2 log 2  (প্রমাণিত)

Koshe dekhi 21 class 9

 7. যদি  ${\color{Blue}&space;\frac{logx}{y-z}=\frac{logy}{z-x}=\frac{logz}{x-y}}$  হয়, তাহলে দেখাই যে xyz = 1 সমাধানঃ  ধরি, $\frac{log\,&space;x}{y-z}=\frac{log\,&space;y}{z-x}=\frac{log\,&space;z}{x-y}=k$ ∴ log x = k (y − z), log y = k (z − x) এবং  log z = k (x − y) এখন, log x + log y + log z = k (y − z) + k (z − x) + k (x − y) বা, log (xyz) = k (y − z + z − x + x − y) বা, log (xyz) = k × 0 বা, log (xyz) = 0  বা, log (xyz) = log 1 ∴ xyz = 1 (প্রমাণিত)

Koshe dekhi 21 class 9

 8. (a) যদি  ${\color{Blue}&space;\frac{log\,&space;x}{b-c}=\frac{log\,&space;y}{c-a}=\frac{log\,&space;z}{a-b}}$   হয়, তাহলে প্রমাণ করি যে  xb + c . yc + a . za + b = 1 সমাধানঃ  ধরি,  $\frac{log\,&space;x}{b-c}=\frac{log\,&space;y}{c-a}=\frac{log\,&space;z}{a-b}=k$ ∴ log x = k (b − c) উভয়দিকে (b + c) গুণ করে পাই, বা, (b + c)log x = k (b − c) (b + c) বা, log x(b + c) = k (b2 − c2) অনুরূপ ভাবে, log y(c + a) = k (c2 − a2) এবং  log z(a + b) = k (a2 − b2) এখন, log x(b + c) + log y(c + a) + log z(a + b) = k (b2 − c2) + k (c2 − a2) + k (a2 − b2) বা, log x(b + c) × y(c + a) × z(a + b) = k (b2 − c2 + c2 − a2 + a2 − b2) বা, log x(b + c) × y(c + a) × z(a + b) = k × 0 বা, log x(b + c) . y(c + a) . z(a + b) = 0 বা, log x(b + c) . y(c + a) . z(a + b) = log 1 ∴ xb + c . yc + a . za + b = 1 (প্রমাণিত)

 8.(b) যদি  ${\color{Blue}&space;\frac{logx}{b-c}=\frac{logy}{c-a}=\frac{logz}{a-b}}$  হয়, তাহলে প্রমাণ করি যে  ${\color{Blue}&space;x^{b^{2}+bc+c^{2}}.y^{c^{2}+ca+a^{2}}.z^{a^{2}+ab+b^{2}}=1}$ সমাধানঃ ধরি,  $\frac{log\,&space;x}{b-c}=\frac{log\,&space;y}{c-a}=\frac{log\,&space;z}{a-b}=k$ ∴ log x = k (b − c) উভয়দিকে (b2 + bc + c2) গুণ করে পাই, বা, (b2 + bc + c2)log x = k (b − c) (b2 + bc + c2) বা, $log\,&space;x^{\left&space;(&space;b^{2}+bc+c^{2}&space;\right&space;)}=k\left&space;(&space;b^{3}-c^{3}&space;\right&space;)$ অনুরূপ ভাবে, $log\,&space;y^{\left&space;(&space;c^{2}+ca+a^{2}&space;\right&space;)}=k\left&space;(&space;c^{3}-a^{3}&space;\right&space;)$ এবং  $log\,&space;z^{\left&space;(&space;a^{2}+ab+b^{2}&space;\right&space;)}=k\left&space;(&space;a^{3}-b^{3}&space;\right&space;)$ এখন, $log\,&space;x^{\left&space;(&space;b^{2}+bc+c^{2}&space;\right&space;)}+log\,&space;y^{\left&space;(&space;c^{2}+ca+a^{2}&space;\right&space;)}+log\,&space;z^{\left&space;(&space;a^{2}+ab+b^{2}&space;\right&space;)}=k\left&space;(&space;b^{3}-c^{3}&space;\right&space;)+k\left&space;(&space;c^{3}-a^{3}&space;\right&space;)+k\left&space;(&space;a^{3}-b^{3}&space;\right&space;)$   বা,$log\,&space;x^{\left&space;(&space;b^{2}+bc+c^{2}&space;\right&space;)}\times&space;y^{\left&space;(&space;c^{2}+ca+a^{2}&space;\right&space;)}\times&space;z^{\left&space;(&space;a^{2}+ab+b^{2}&space;\right&space;)}=k\left&space;(&space;b^{3}-c^{3}+c^{3}-a^{3}+a^{3}-b^{3}&space;\right&space;)$   বা, $log\,&space;x^{\left&space;(&space;b^{2}+bc+c^{2}&space;\right&space;)}.\,&space;y^{\left&space;(&space;c^{2}+ca+a^{2}&space;\right&space;)}.\,&space;z^{\left&space;(&space;a^{2}+ab+b^{2}&space;\right&space;)}=k\times&space;0$ বা, $log\,&space;x^{\left&space;(&space;b^{2}+bc+c^{2}&space;\right&space;)}.\,&space;y^{\left&space;(&space;c^{2}+ca+a^{2}&space;\right&space;)}.\,&space;z^{\left&space;(&space;a^{2}+ab+b^{2}&space;\right&space;)}=0$ বা, $log\,&space;x^{\left&space;(&space;b^{2}+bc+c^{2}&space;\right&space;)}.\,&space;y^{\left&space;(&space;c^{2}+ca+a^{2}&space;\right&space;)}.\,&space;z^{\left&space;(&space;a^{2}+ab+b^{2}&space;\right&space;)}=log\,&space;1$ ${\color{DarkGreen}&space;\therefore&space;x^{b^{2}+bc+c^{2}}.\,&space;y^{c^{2}+ca+a^{2}}.\,&space;z^{a^{2}+ab+b^{2}}=1}$  (প্রমাণিত)

 9. যদি  a3 − x . b5x = a5 + x . b3x  হয়, তাহলে দেখাই যে  ${\color{Blue}&space;xlog\left&space;(&space;\frac{b}{a}&space;\right&space;)=loga}$ সমাধানঃ  a3 − x . b5x = a5 + x . b3x বা, $\frac{b^{5x}}{b^{3x}}=\frac{a^{5+x}}{a^{3-x}}$ বা, b5x − 3x = a5+x − 3+x বা, b2x = a2 + 2x বা, b2x = a2 . a2x বা, $\frac{b^{2x}}{a^{2x}}=a^{2}$ বা, $\left&space;(\frac{b}{a}&space;\right&space;)^{2x}=a^{2}$ a3 − x . b5x = a5 + x . b3x বা, $\frac{b^{5x}}{b^{3x}}=\frac{a^{5+x}}{a^{3-x}}$ বা, b5x − 3x = a5+x − 3+x  ${\color{Blue}&space;\left&space;[&space;\because&space;log\,&space;_{a}\,&space;\frac{m}{n}=log\,&space;_{a}\,&space;m-log\,&space;_{a}\,&space;n&space;\right&space;]}$ বা, b2x = a2 + 2x বা, b2x = a2 . a2x  ${\color{Blue}&space;\left&space;[&space;\because&space;log\,&space;_{a}\,&space;mn=log\,&space;_{a}\,&space;m+log\,&space;_{a}\,&space;n&space;\right&space;]}$ বা, $\frac{b^{2x}}{a^{2x}}=a^{2}$ বা, $\left&space;(\frac{b}{a}&space;\right&space;)^{2x}=a^{2}$ বা, $\left&space;[\left&space;(\frac{b}{a}&space;\right&space;)^{x}&space;\right&space;]^{2}=a^{2}$ বা, $\left&space;(\frac{b}{a}&space;\right&space;)^{x}=a$ উভয়দিকে log নিয়ে পাই, বা, $log\left&space;(\frac{b}{a}&space;\right&space;)^{x}=log\,&space;a$ ${\color{DarkGreen}&space;\therefore&space;x\,&space;log\left&space;(\frac{b}{a}&space;\right&space;)=log\,&space;a}$ ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$  (প্রমাণিত)

 10. সমাধান করি : (a) ${\color{Blue}&space;log_{8}\left&space;[&space;log_{2}\left&space;\{&space;log_{3}\left&space;(&space;4^{x}+17&space;\right&space;)&space;\right&space;\}&space;\right&space;]=\frac{1}{3}}$ সমাধানঃ  $log_{8}\left&space;[&space;log_{2}\left&space;\{&space;log_{3}\left&space;(&space;4^{x}+17&space;\right&space;)&space;\right&space;\}&space;\right&space;]=\frac{1}{3}$ বা, $\left&space;[&space;log_{2}\left&space;\{&space;log_{3}\left&space;(&space;4^{x}+17&space;\right&space;)&space;\right&space;\}&space;\right&space;]=8^{\frac{1}{3}}$  [যেহেতু, loga b = c হলে, ac = b হয়] বা, $\left&space;[&space;log_{2}\left&space;\{&space;log_{3}\left&space;(&space;4^{x}+17&space;\right&space;)&space;\right&space;\}&space;\right&space;]=\left&space;(2^{3}&space;\right&space;)^{\frac{1}{3}}$ বা, $\left&space;[&space;log_{2}\left&space;\{&space;log_{3}\left&space;(&space;4^{x}+17&space;\right&space;)&space;\right&space;\}&space;\right&space;]=2$ বা, $\left&space;\{&space;log_{3}\left&space;(&space;4^{x}+17&space;\right&space;)&space;\right&space;\}=2^{2}$  [যেহেতু, loga b = c হলে, ac = b হয়] বা, $\left&space;\{&space;log_{3}\left&space;(&space;4^{x}+17&space;\right&space;)&space;\right&space;\}=4$ বা, $\left&space;(&space;4^{x}+17&space;\right&space;)=3^{4}$  [যেহেতু, loga b = c হলে, ac = b হয়] বা, 4x + 17 = 81 বা, 4x = 81 − 17 বা, 4x = 64 বা, 4x = 43 ∴ x = 3  (উত্তর)

 10. সমাধান করি : (b) log8 x + log4 x + log2 x = 11 সমাধানঃ  log8 x + log4 x + log2 x = 11 বা, $\frac{log\,&space;x}{log\,&space;8}+\frac{log\,&space;x}{log\,&space;4}+\frac{log\,&space;x}{log\,&space;2}=11$ বা, $\frac{log\,&space;x}{log\,&space;2^{3}}+\frac{log\,&space;x}{log\,&space;2^{2}}+\frac{log\,&space;x}{log\,&space;2}=11$ বা, $\frac{log\,&space;x}{3log\,&space;2}+\frac{log\,&space;x}{2log\,&space;2}+\frac{log\,&space;x}{log\,&space;2}=11$ বা, $\frac{log\,&space;x}{log\,&space;2}\times&space;\left&space;(&space;\frac{1}{3}+\frac{1}{2}+1&space;\right&space;)=11$ বা, $\frac{log\,&space;x}{log\,&space;2}\times&space;\left&space;(&space;\frac{2+3+6}{6}&space;\right&space;)=11$ বা, $\frac{log\,&space;x}{log\,&space;2}\times&space;\left&space;(&space;\frac{11}{6}&space;\right&space;)=11$ বা, $\frac{log\,&space;x}{log\,&space;2}=11\times&space;\frac{6}{11}$ বা, $\frac{log\,&space;x}{log\,&space;2}=6$ বা, log x = 6 log 2 বা, log x =  log 26  ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ বা, x = 26 ∴ x = 64 (উত্তর)

 11. দেখাই log10 2  -এর মান ${\color{Blue}&space;\frac{1}{4}}$ এবং ${\color{Blue}&space;\frac{1}{3}}$ -এর মধ্যে অবস্থিত।  সমাধানঃ  ধরি, x = log10 2 ∴ 10x = 2 [যেহেতু, loga b = c হলে, ac = b হয়] বা, 1012x = 212 বা, 1012x = 4096 এখন, 1000 < 4096 < 10000 = 103 < 4096 < 104 = 103 < 1012x < 104 = 3 < 12x < 4 $=\frac{3}{12}<\frac{12x}{12}<\frac{4}{12}$ $=\frac{1}{4} $=\frac{1}{4} (প্রমাণিত)

Koshe dekhi 21 class 9

 12. বহু বিকল্পীয় প্রশ্ন (M.C.Q) (i)  যদি log√x 0.25 = 4 হয়, তাহলে x -এর মান  (a) 0.5  (b) 0.25  (c) 4  (d) 16 সমাধানঃ  log√x 0.25 = 4 বা, (√x)4 = 0.25  [যেহেতু, loga b = c হলে, ac = b হয়] বা, x2 = (0.5)2 ∴ x = 0.5 উত্তরঃ  (a) 0.5

Koshe dekhi 21 class 9

 12. বহু বিকল্পীয় প্রশ্ন (M.C.Q) (ii) log10 (7x − 5) = 2 হলে, x -এর মান (a) 10  (b) 12  (c) 15  (d) 18 সমাধানঃ  log10 (7x − 5) = 2 বা, 102 = (7x − 5)  [যেহেতু, loga b = c হলে, ac = b হয়] বা, 100 = (7x − 5) বা, 100 + 5 = 7x বা, 105 = 7x বা, $x=\frac{105}{7}$ ∴ x = 15 উত্তরঃ  (c) 15

Koshe dekhi 21 class 9

 12. বহু বিকল্পীয় প্রশ্ন (M.C.Q)  (iii) log2 3 = a হলে, log8 27 হবে   (a) 3a  (b) ${\color{Blue}&space;\frac{1}{a}}$  (c) 2a  (d) a সমাধানঃ  log8 27 $=\frac{log\,&space;27}{log\,&space;8}$ $=\frac{log\,&space;3^{3}}{log\,&space;2^{3}}$ $=\frac{3log\,&space;3}{3log\,&space;2}$ $=\frac{log\,&space;3}{log\,&space;2}$ = log2 3 = a উত্তরঃ (d) a

Koshe dekhi 21 class 9

 12. বহু বিকল্পীয় প্রশ্ন (M.C.Q) (iv) log√2 x = a হলে, log2√2 x হবে   (a) ${\color{Blue}&space;\frac{a}{3}}$  (b) a  (c) 2a  (d) 3a সমাধানঃ  log2√2 x $=\frac{log\,&space;x}{log\,&space;2\sqrt{2}}$ $=\frac{log\,&space;x}{log\,&space;\left&space;(\sqrt{2}&space;\right&space;)^{3}}$ $=\frac{log\,&space;x}{3log\,&space;\sqrt{2}}$ $=\frac{1}{3}\times&space;log\,&space;_{\sqrt{2}}\,&space;x$ $=\frac{1}{3}\times&space;a$ $=\frac{a}{3}$ উত্তরঃ  (a) ${\color{DarkGreen}&space;\frac{a}{3}}$

Koshe dekhi 21 class 9

 12. বহু বিকল্পীয় প্রশ্ন (M.C.Q) (v) ${\color{Blue}&space;log_{x}\frac{1}{3}=-\frac{1}{3}}$  হলে, x -এর মান হবে   (a) 27  (b) 9  (c) 3  (d) ${\color{Blue}&space;\frac{1}{27}}$ সমাধানঃ  $log_{x}\,&space;\frac{1}{3}=-\frac{1}{3}$  বা,  ${x}^{-\frac{1}{3}}=\frac{1}{3}$   [যেহেতু, loga b = c হলে, ac = b হয়]  বা, ${x}^{-\frac{1}{3}}=3^{-1}$  বা, ${x}^{-\frac{1}{3}}=\left&space;(3^{3}&space;\right&space;)^{-\frac{1}{3}}$  বা, x = 33 ∴ x = 27 উত্তরঃ  (a) 27

Koshe dekhi 21 class 9

 13. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্ন : (i) log4 log4 log4 256 -এর মান কত হবে হিসাব করি।  সমাধানঃ  log4 log4 log4 256 $=log_{4}\,&space;log_{4}\,&space;log_{4}\,&space;4^{4}$ $=log_{4}\,&space;log_{4}\,&space;4log_{4}\,&space;4$ ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ $=log_{4}\,&space;log_{4}\,&space;4\times&space;1$ ${\color{Blue}&space;[\because&space;log_{a}\,&space;a=1]}$ $=log_{4}\,&space;log_{4}\,&space;4$ $=log_{4}\,&space;1$ ${\color{Blue}&space;[\because&space;log_{a}\,&space;a=1]}$ = 0 (উত্তর)

Koshe dekhi 21 class 9

 13. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্ন : (ii) ${\color{Blue}&space;log\frac{a^{n}}{b^{n}}+log\frac{b^{n}}{c^{n}}+log\frac{c^{n}}{a^{n}}}$ -এর মান কত হবে হিসাব করি।  সমাধানঃ $log\frac{a^{n}}{b^{n}}+log\frac{b^{n}}{c^{n}}+log\frac{c^{n}}{a^{n}}$ = log an − log bn + log bn − log cn + log cn − log an = 0 (উত্তর)

Koshe dekhi 21 class 9

 13. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্ন : (iii) দেখাই যে ${\color{Blue}&space;a^{log_{a}x}=x}$ সমাধানঃ  ধরি, $k=a^{log_{a}x}$ উভয়পক্ষে log নিয়ে পাই, $log\,&space;k=log\,&space;a^{log_{a}x}$ বা, $log\,&space;k=log\,&space;a\times&space;log_{a}x$ বা, $log\,&space;k=log\,&space;a\times&space;\frac{log\,&space;x}{log\,&space;a}$ বা, log k = log x বা, k = x ${\color{DarkGreen}&space;\therefore&space;a^{log_{a}x}=x}$ (প্রমাণিত)

Koshe dekhi 21 class 9

 13. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্ন : (iv) loge 2. logx 25 = log10 16. loge 10 হলে, x -এর মান নির্ণয় করি।  সমাধানঃ  loge 2. logx 25 = log10 16. loge 10 বা, $log_{e}\,&space;2.log_{x}\,&space;25=\frac{log\,&space;16}{log\,&space;10}\times&space;\frac{log\,&space;10}{log\,&space;e}$ বা, $log_{e}\,&space;2.log_{x}\,&space;25=log_{e}\,&space;16$ বা, $log_{e}\,&space;2.log_{x}\,&space;25=log_{e}\,&space;2^{4}$ বা, $log_{e}\,&space;2.log_{x}\,&space;25=4\,&space;log_{e}\,&space;2$ ${\color{Blue}&space;[\because&space;log\,&space;_{a}\,&space;m^{c}=c\,&space;log\,&space;_{a}\,&space;m]}$ বা, $log_{x}\,&space;25=\frac{4\,&space;log_{e}\,&space;2}{log_{e}\,&space;2}$ বা, $log_{x}\,&space;25=4$ বা, x4 = 25 বা, x4 = (√5)4 ∴ x = √5  (উত্তর)

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#### Koshe dekhi 20.3 Class 8

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