Koshe dekhi 21 class 9

1. মান নির্ণয় করি :

(i) ${\color{Blue} log_{4}\left ( \frac{1}{64} \right )}$

সমাধানঃ

$log_{4}\left ( \frac{1}{64} \right )$

$=-log_{4}\, 64\; {\color{Blue} [\because log\left ( \frac{1}{a} \right )=-log\, a]}$

$=-log_{4}\, 4^{3}$

$=-3\, log_{4}\, 4\; {\color{Blue} [\because log\, a^{m}=mlog\, a]}$

= −3 × (1) ${\color{Blue} [\because log_{a}\, a=1]}$

= −3 (উত্তর)

1. মান নির্ণয় করি :

(ii) ${\color{Blue} log_{0.01}0.000001}$

সমাধানঃ

$log_{0.01}0.000001$

$=\frac{log\, 0.000001}{log\, 0.01}$ ${\color{Blue} [\because log_{a}\, b=\frac{logb}{loga}]}$

$=\frac{log\, \left ( \frac{1}{1000000} \right )}{log\, \left (\frac{1}{100} \right )}$

$=\frac{-log\, 100000}{-log\, 100}\; {\color{Blue} [\because log\, \left ( \frac{1}{a} \right )=-log\, a]}$

$=\frac{log\, 10^{6}}{log\, 10^{2}}$

$=\frac{6\, log\, 10}{2\, log\, 10}\; {\color{Blue} [\because log\, a^{m}=m\, log\, a]}$

= 3 (উত্তর)

অথবা

$log_{0.01}0.000001$

$=log_{0.01}\left ( 0.01 \right )^{3}$

= 3 log_0.01 0.01 ${\color{Blue} [\because log\, a^{m}=m\, log\, a]}$

= 3 ${\color{Blue} [\because log_{a}\, a=1]}$ (উত্তর)

1. মান নির্ণয় করি :

(iii) ${\color{Blue} log_{\sqrt{6}}216}$

সমাধানঃ

$log_{\sqrt{6}}216$

$=\frac{log\, 216}{log\, \sqrt{6}}$ ${\color{Blue} [\because log_{a}\, b=\frac{logb}{loga}]}$

$=\frac{log\, 6^{3}}{log\, 6^{\frac{1}{2}}}$

$=\frac{3\, log\, 6}{\frac{1}{2}\, log\, 6}$ ${\color{Blue} [\because log\, a^{m}=m\, log\, a]}$

$=\frac{3}{\left (\frac{1}{2} \right )}$

$=3\times \frac{2}{1}$

= 6 (উত্তর)

1. মান নির্ণয় করি :

(iv) ${\color{Blue} log_{2\sqrt{3}}1728}$

সমাধানঃ

$log_{2\sqrt{3}}1728$

$=\frac{log\, 1728}{log\, 2\sqrt{3}}$ ${\color{Blue} [\because log_{a}\, b=\frac{logb}{loga}]}$

∴ 1728

= 2⁶ × 3³

= 2⁶ × (√3)⁶

= (2√3)⁶

এখন, $\frac{log\, 1728}{log\, 2\sqrt{3}}$

$=\frac{log\, \left ( 2\sqrt{3} \right )^{6}}{log\, \left ( 2\sqrt{3} \right )}$

$=\frac{6\, log\, 2\sqrt{3}}{log\, 2\sqrt{3}}$

= 6 (উত্তর)

2. (a) 625 -এর লগারিদম 4 হলে, নিধান কি হবে হিসাব করে লিখি।

সমাধানঃ

ধরি, নিধান হল ‘x‘

∴ log_x 625 = 4

বা, x⁴ = 625 [যেহেতু, log_a b = c হলে, a^c = 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‘

∴ log_x 5832 = 6

বা, x⁶ = 5832 [যেহেতু, log_a b = c হলে, a^c = 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) ${\color{Blue} 1+log_{10}\, a=2\, log_{10}\, b}$ হলে, a কে b -এর দ্বারা প্রকাশ করি।

সমাধানঃ

$1+log_{10}\, a=2\, log_{10}\, b$

বা, 1 = 2 log₁₀ b − log₁₀ a

বা, log₁₀ 10 = log₁₀ b² − log₁₀ a ${\color{Blue} [\because log_{a}\, a=1]}$ এবং ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

বা, $log\, _{10}\, 10=log\, _{10}\,\frac{b^{2}}{a}$ ${\color{Blue} \left [ \because log\, _{a}\, m-log\, _{a}\, n=log\, _{a}\, \frac{m}{n} \right ]}$

বা, $10=\frac{b^{2}}{a}$

${\color{DarkGreen} \therefore a=\frac{b^{2}}{10}}$ (উত্তর)

3. (b) ${\color{Blue} 3+log\, _{10}\, x=2\, log\, _{10}\, y}$ হলে, x কে y -এর দ্বারা প্রকাশ করি।

সমাধানঃ

$3+log\, _{10}\, x=2\, log\, _{10}\, y$

বা, $log\, _{10}\, x=2\, log\, _{10}\, y-3$

বা, $log\, _{10}\, x=2\, log\, _{10}\, y-3$ × 1

বা, $log\, _{10}\, x=2\, log\, _{10}\, y-3\, log\, _{10}\, 10$ ${\color{Blue} [\because log_{a}\, a=1]}$

বা, $log\, _{10}\, x=log\, _{10}\, y^{2}-log\, _{10}\, 10^{3}$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

বা, $log\, _{10}\, x=log\, _{10}\,\frac{y^{2}}{10^{3}}$ ${\color{Blue} \left [ \because log\, _{a}\, m-log\, _{a}\, n=log\, _{a}\, \frac{m}{n} \right ]}$

${\color{DarkGreen} \therefore x=\frac{y^{2}}{1000}}$ (উত্তর)

4. মান নির্ণয় করি :

(a) ${\color{Blue} log_{2}\left [ log_{2}\left \{ log_{3}\left ( log_{3}27^{3} \right ) \right \} \right ]}$

সমাধানঃ

$log_{2}\left [ log_{2}\left \{ log_{3}\left ( log_{3}\, 27^{3} \right ) \right \} \right ]$

$=log_{2}\left [ log_{2}\left \{ log_{3}\left ( log_{3}\, 3^{3^{3}} \right ) \right \} \right ]$

$=log_{2}\left [ log_{2}\left \{ log_{3}\left ( log_{3}\, 3^{9} \right ) \right \} \right ]$

$=log_{2}\left [ log_{2}\left \{ log_{3}\left ( 9\, log_{3}\, 3 \right ) \right \} \right ]$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

$=log_{2}\left [ log_{2}\left \{ log_{3}\left ( 9\times 1 \right ) \right \} \right ]$ ${\color{Blue} [\because log_{a}\, a=1]}$

$=log_{2}\left [ log_{2}\left \{ log_{3}\, 9 \right \} \right ]$

$=log_{2}\left [ log_{2}\left \{ log_{3}\, 3^{2} \right \} \right ]$

$=log_{2}\left [ log_{2}\left \{ 2\, log_{3}\, 3 \right \} \right ]$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

$=log_{2}\left [ log_{2}\left \{ 2\times 1\right \} \right ]$ ${\color{Blue} [\because log_{a}\, a=1]}$

$=log_{2}\left [ log_{2}\, 2 \right ]$

$=log_{2}\, 1$

= 0 ${\color{Blue} \left [\because log_{x}\, 1=0 \right ]}$ (উত্তর)

4. মান নির্ণয় করি :

(b) ${\color{Blue} \frac{log\sqrt{27}+log8-log\sqrt{1000}}{log1.2}}$

সমাধানঃ

$\frac{log\sqrt{27}+log8-log\sqrt{1000}}{log1.2}$

$=\frac{log\, \sqrt{3^{3}}+log\, 2^{3}-log\, \sqrt{10^{3}}}{log\, \frac{12}{10}}$

$=\frac{log\, 3^{\frac{3}{2}}+3\, log\, 2-log\, 10^{\frac{3}{2}}}{log\, \frac{3\times 4}{10}}$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

$=\frac{\frac{3}{2}\, log\, 3+3\, log\, 2-\frac{3}{2}\, log\, 10}{log\, 3+log\, 4-log\, 10}$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$ , ${\color{Blue} \left [ \because log\, _{a}\, mn=log\, _{a}\, m+log\, _{a}\, n \right ]}$ এবং ${\color{Blue} \left [ \because log\, _{a}\, \frac{m}{n}=log\, _{a}\, m-log\, _{a}\, n \right ]}$

$=\frac{\frac{3}{2}\, log\, 3+\frac{3}{2}\times 2\, log\, 2-\frac{3}{2}\, log\, 10}{log\, 3+log\, 2^{2}-log\, 10}$

$=\frac{\frac{3}{2}\left (log\, 3+2\, log\, 2-log\, 10 \right )}{\left (log\, 3+2\, log\, 2-log\, 10 \right )}$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

${\color{DarkGreen} =\frac{3}{2}}$ (উত্তর)

4. মান নির্ণয় করি :

সমাধানঃ

$log_{3}\, 4\times log_{4}\, 5\times log_{5}\, 6\times log_{6}\, 7\times log_{7}\, 3$

$=\frac{log\, 4}{log\, 3}\times \frac{log\, 5}{log\, 4}\times \frac{log\, 6}{log\, 5}\times \frac{log\, 7}{log\, 6}\times \frac{log\, 3}{log\, 7}$

[ধরি, প্রতিটির log এর নিধান একই]

= 1 (উত্তর)

4. মান নির্ণয় করি :

(d) ${\color{Blue} log_{10}\frac{384}{5}+log_{10}\frac{81}{32}+3log_{10}\frac{5}{3}+log_{10}\frac{1}{9}}$

সমাধানঃ

$log_{10}\, \frac{384}{5}+log_{10}\, \frac{81}{32}+3log_{10}\, \frac{5}{3}+log_{10}\, \frac{1}{9}$

$=log_{10}\, 384-log_{10}\, 5+log_{10}\, 81-log_{10}\, 32+3log_{10}\, 5-3log_{10}\, 3+log_{10}\, 1-log_{10}\, 9$

$=log_{10}\, 2^{7}\times 3-log_{10}\, 5+log_{10}\, 3^{4}-log_{10}\, 2^{5}+3log_{10}\, 5-3log_{10}\, 3+log_{10}\, 1-log_{10}\, 3^{2}$

$=7log_{10}\, 2+log_{10}\, 3-log_{10}\, 5+4log_{10}\, 3-5log_{10}\, 2+3log_{10}\, 5-3log_{10}\, 3+log_{10}\, 1-2log_{10}\, 3$

$=2log_{10}\, 2+2log_{10}\, 5$ ${\color{Blue} \left [\because log_{x}\, 1=0 \right ]}$

$=log_{10}\, 2^{2}+log_{10}\, 5^{2}$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

$=log_{10}\, 4+log_{10}\, 25$

$=log_{10}\, \left (4\times 25 \right )$ ${\color{Blue} \left [ \because log\, _{a}\, mn=log\, _{a}\, m+log\, _{a}\, n \right ]}$

$=log_{10}\, 100$

$=log_{10}\, 10^{2}$

$=2log_{10}\, 10$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

= 2 (উত্তর)

5. প্রমাণ করি :

(i) ${\color{Blue} log\frac{75}{16}-2log\frac{5}{9}+log\frac{32}{243}=log2}$

সমাধানঃ

বামপক্ষ :

$log\frac{75}{16}-2log\frac{5}{9}+log\frac{32}{243}$

$=log\, 75-log\, 16-2log\, 5+2log\, 9+log\, 32-log\, 243$

$=log\, \left (5^{2}\times 3 \right )-log\, 2^{4}-2log\, 5+2log\, 3^{2}+log\, 2^{5}-log\, 3^{5}$

$=log\, 5^{2}+log\, 3-log\, 2^{4}-2log\, 5+2\times 2log\, 3+log\, 2^{5}-log\, 3^{5}$

${\color{Blue} \left [ \because log\, _{a}\, mn=log\, _{a}\, m+log\, _{a}\, n \right ]}$

$=2log\, 5+log\, 3-4log\, 2-2log\, 5+4log\, 3+5log\, 2-5log\, 3$

${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

= log 2

= ডানপক্ষ (প্রমাণিত)

5. প্রমাণ করি :

(ii) ${\color{Blue} log_{10}15\left ( 1+log_{15}30 \right )+\frac{1}{2}log_{10}16\left ( 1+log_{4}7\right )-log_{10}6\left ( log_{6}3+1+log_{6}7\right )=2}$

সমাধানঃ

বামপক্ষ :

$log_{10}15\left ( 1+log_{15}30 \right )+\frac{1}{2}log_{10}16\left ( 1+log_{4}7\right )-log_{10}6\left ( log_{6}3+1+log_{6}7\right )$

$=log_{10}\, 15+log_{10}\, 15\times log_{15}\, 30 +\frac{1}{2}log_{10}\, 16+\frac{1}{2}log_{10}\, 16\times log_{4}\, 7-log_{10}\, 6\times log_{6}\, 3-log_{10}\, 6-log_{10}\, 6\times log_{6}\, 7$

$=log_{10}\, 15+\left (\frac{log\, 15}{log\, 10}\times \frac{log\, 30}{log\, 15} \right )+\frac{1}{2}log_{10}\, 4^{2}+\frac{1}{2}log_{10}\, 4^{2}\times log_{4}\, 7-\left (\frac{log\, 6}{log\, 10}\times \frac{log\, 3}{log\, 6} \right )-log_{10}\, 6-\left (\frac{log\, 6}{log\, 10}\times \frac{log\, 7}{log\, 6} \right )$

$=log_{10}\, 15+log_{10}\, 30 +\frac{1}{2}\times 2\, log_{10}\, 4+\frac{1}{2}\times 2\, log_{10}\, 4\times log_{4}\, 7-log_{10}\, 3-log_{10}\, 6-log_{10}\, 7$

$=log_{10}\, 15+log_{10}\, 30 +log_{10}\, 4+\left ( \frac{log\, 4}{log\, 10}\times \frac{log\, 7}{log\, 4} \right )-log_{10}\, 3-log_{10}\, 6-log_{10}\, 7$

$=log_{10}\, 15+log_{10}\, 30 +log_{10}\, 4+log_{10}\, 7-log_{10}\, 3-log_{10}\, 6-log_{10}\, 7$

$=\left (log_{10}\, 15+log_{10}\, 30 +log_{10}\, 4+log_{10}\, 7 \right )-\left (log_{10}\, 3+log_{10}\, 6+log_{10}\, 7 \right )$

$=log_{10}\, \frac{15\times 30\times 4\times 7}{3\times 6\times 7}$

$=log_{10}\, 100$

$=log_{10}\, 10^{2}$

$=2log_{10}\, 10$

= 2

= ডানপক্ষ (প্রমাণিত)

5. প্রমাণ করি :

(iii) ${\color{Blue} log_{2}log_{2}log_{4}\, 256+2log_{\sqrt{2}}\, 2=5}$

সমাধানঃ

বামপক্ষ :

$log_{2}log_{2}log_{4}\, 256+2log_{\sqrt{2}}\, 2$

$=log_{2}log_{2}log_{4}\, 4^{4}+2log_{\sqrt{2}}\, \left (\sqrt{2} \right )^{2}$

$=log_{2}\, log_{2}\, 4log_{4}\, 4+2\times 2\, log_{\sqrt{2}}\, \sqrt{2}$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

$=log_{2}\, log_{2}\, 4\times 1+4\times 1$ ${\color{Blue} [\because log_{a}\, a=1]}$

$=log_{2}\, log_{2}\, 4+4$

$=log_{2}\, log_{2}\, 2^{2}+4$

$=log_{2}\, 2log_{2}\, 2+4$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

$=log_{2}\, 2\times 1+4$ ${\color{Blue} [\because log_{a}\, a=1]}$

$=log_{2}\, 2+4$

= 1 + 4 ${\color{Blue} [\because log_{a}\, a=1]}$

= 5

= ডানপক্ষ (প্রমাণিত)

5. প্রমাণ করি :

(iv) ${\color{Blue} log_{x^{2}}\, x\times log_{y^{2}}\, y\times log_{z^{2}}\, z=\frac{1}{8}}$

সমাধানঃ

বামপক্ষ :

$log_{x^{2}}\, x\times log_{y^{2}}\, y\times log_{z^{2}}\, z$

$=log_{x^{2}}\, \left (x^{2} \right )^{\frac{1}{2}}\times log_{y^{2}}\, \left (y^{2} \right )^{\frac{1}{2}}\times log_{z^{2}}\, \left (z^{2} \right )^{\frac{1}{2}}$

$=\frac{1}{2}\, log_{x^{2}}\, x^{2}\times \frac{1}{2}\, log_{y^{2}}\, y^{2}\times \frac{1}{2}\, log_{z^{2}}\, z^{2}$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

$=\frac{1}{2}\times 1\times \frac{1}{2}\times 1\times \frac{1}{2}\times 1$ ${\color{Blue} [\because log_{a}\, a=1]}$

$=\frac{1}{8}$

= ডানপক্ষ (প্রমাণিত)

5. প্রমাণ করি :

(v) ${\color{Blue} log_{b^{3}}\, a\times log_{c^{3}}\, b\times log_{a^{3}}\, c=\frac{1}{27}}$

সমাধানঃ

বামপক্ষ :

$log_{b^{3}}\, a\times log_{c^{3}}\, b\times log_{a^{3}}\, c$

$=\frac{log\, a}{log\, b^{3}}\times \frac{log\, b}{log\, c^{3}}\times \frac{log\, c}{log\, a^{3}}$

$=\frac{log\, a}{3\, log\, b}\times \frac{log\, b}{3\, log\, c}\times \frac{log\, c}{3\, log\, a}$

$=\frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}$

$=\frac{1}{27}$

= ডানপক্ষ (প্রমাণিত)

Koshe dekhi 21 class 9

5. প্রমাণ করি :

(vi) ${\color{Blue} \frac{1}{log_{xy}\left ( xyz \right )}+\frac{1}{log_{yz}\left ( xyz \right )}+\frac{1}{log_{zx}\left ( xyz \right )}=2}$

সমাধানঃ

বামপক্ষ :

$\frac{1}{log_{xy}\left ( xyz \right )}+\frac{1}{log_{yz}\left ( xyz \right )}+\frac{1}{log_{zx}\left ( xyz \right )}$

$=\frac{log\, xy}{log\, xyz}+\frac{log\, yz}{log\, xyz}+\frac{log\, zx}{log\, xyz}$

$=\frac{log\, xy+log\, yz+log\, zx}{log\, xyz}$

$=\frac{log\, \left (xy\times yz\times zx \right )}{log\, xyz}$ ${\color{Blue} \left [ \because log\, _{a}\, mn=log\, _{a}\, m+log\, _{a}\, n \right ]}$

$=\frac{log\, \left (x^{2}y^{2}z^{2} \right )}{log\, xyz}$

$=\frac{log\, \left (xyz \right )^{2}}{log\, xyz}$

$=\frac{2\, log\, xyz}{log\, xyz}$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

= 2

= ডানপক্ষ (প্রমাণিত)

Koshe dekhi 21 class 9

5. প্রমাণ করি :

(vii) ${\color{Blue} log\frac{a^{2}}{bc}+log\frac{b^{2}}{ca}+log\frac{c^{2}}{ab}=0}$

সমাধানঃ

বামপক্ষ :

$log\frac{a^{2}}{bc}+log\frac{b^{2}}{ca}+log\frac{c^{2}}{ab}$

$=log\, a^{2}-log\, bc+log\, b^{2}-log\, ca+log\, c^{2}-log\, ab$

$=\left (log\, a^{2}+log\, b^{2}+log\, c^{2} \right )-\left (log\, bc+log\, ca+log\, ab \right )$

$=log\, \left (a^{2}\times b^{2}\times c^{2} \right )-log\, \left ( bc\times ca\times ab \right )$

$=log\left ( a^{2}b^{2}c^{2} \right )-log\left ( a^{2}b^{2}c^{2} \right )$

= 0

= ডানপক্ষ (প্রমাণিত)

Koshe dekhi 21 class 9

5. প্রমাণ করি :

(viii) x^{log y − log z}× y^{log z − log x}× z^{log x − log y = 1}

সমাধানঃ

ধরি,

k = x^{log y − log z}× y^{log z − log x}× z^{log x − log y}

উভয়পক্ষে log নিয়ে পাই,

log k = log (x^{log y − log z}× y^{log z − log x}× z^{log 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

∴ x^{log y − log z}× y^{log z − log x}× z^{log x − log y = 1}

বামপক্ষ = ডানপক্ষ (প্রমাণিত)

Koshe dekhi 21 class 9

6.(i) যদি ${\color{Blue} log\, \frac{x+y}{5}=\frac{1}{2}\left ( log\, x+log\, y \right )}$ হয়, তাহলে দেখাই যে ${\color{Blue} \frac{x}{y}+\frac{y}{x}=23}$

সমাধানঃ

$log\, \frac{x+y}{5}=\frac{1}{2}\left ( log\, x+log\, y \right )$

বা, $2\, log\, \frac{x+y}{5}=\left ( log\, x+log\, y \right )$

বা, $log\, \left (\frac{x+y}{5} \right )^{2}=log\, xy$

বা, $\left (\frac{x+y}{5} \right )^{2}=xy$

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, $\frac{x^{2}}{xy}+\frac{y^{2}}{xy}=23$

${\color{DarkGreen} \therefore \frac{x}{y}+\frac{y}{x}=23}$ (প্রমাণিত)

Koshe dekhi 21 class 9

6(ii) যদি a⁴+ b⁴= 14a²b² হয়, তাহলে দেখাই যে log (a²+ b²) = log a + log b + 2 log 2

সমাধানঃ

a⁴+ b⁴= 14a²b²

বা, (a²+ b²)²− 2a²b² = 14a²b²

বা, (a²+ b²)²= 14a²b² + 2a²b²

বা, (a²+ b²)²= 16a²b²

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 ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

∴ log (a²+ b²) = log a + log b + 2 log 2 (প্রমাণিত)

Koshe dekhi 21 class 9

7. যদি ${\color{Blue} \frac{logx}{y-z}=\frac{logy}{z-x}=\frac{logz}{x-y}}$ হয়, তাহলে দেখাই যে xyz = 1

সমাধানঃ

ধরি, $\frac{log\, x}{y-z}=\frac{log\, y}{z-x}=\frac{log\, z}{x-y}=k$

∴ 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 ${\color{Blue} \frac{log\, x}{b-c}=\frac{log\, y}{c-a}=\frac{log\, z}{a-b}}$ হয়, তাহলে প্রমাণ করি যে

x^{b + c}. y^{c + a}. z^{a + b}= 1

সমাধানঃ

ধরি, $\frac{log\, x}{b-c}=\frac{log\, y}{c-a}=\frac{log\, z}{a-b}=k$

∴ 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

∴ x^{b + c}. y^{c + a}. z^{a + b}= 1 (প্রমাণিত)

8.(b) যদি ${\color{Blue} \frac{logx}{b-c}=\frac{logy}{c-a}=\frac{logz}{a-b}}$ হয়, তাহলে প্রমাণ করি যে

${\color{Blue} x^{b^{2}+bc+c^{2}}.y^{c^{2}+ca+a^{2}}.z^{a^{2}+ab+b^{2}}=1}$

সমাধানঃ

ধরি, $\frac{log\, x}{b-c}=\frac{log\, y}{c-a}=\frac{log\, z}{a-b}=k$

∴ 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, $log\, x^{\left ( b^{2}+bc+c^{2} \right )}=k\left ( b^{3}-c^{3} \right )$

অনুরূপ ভাবে,

$log\, y^{\left ( c^{2}+ca+a^{2} \right )}=k\left ( c^{3}-a^{3} \right )$ এবং

$log\, z^{\left ( a^{2}+ab+b^{2} \right )}=k\left ( a^{3}-b^{3} \right )$

এখন,

$log\, x^{\left ( b^{2}+bc+c^{2} \right )}+log\, y^{\left ( c^{2}+ca+a^{2} \right )}+log\, z^{\left ( a^{2}+ab+b^{2} \right )}=k\left ( b^{3}-c^{3} \right )+k\left ( c^{3}-a^{3} \right )+k\left ( a^{3}-b^{3} \right )$

বা, $log\, x^{\left ( b^{2}+bc+c^{2} \right )}\times y^{\left ( c^{2}+ca+a^{2} \right )}\times z^{\left ( a^{2}+ab+b^{2} \right )}=k\left ( b^{3}-c^{3}+c^{3}-a^{3}+a^{3}-b^{3} \right )$

বা, $log\, x^{\left ( b^{2}+bc+c^{2} \right )}.\, y^{\left ( c^{2}+ca+a^{2} \right )}.\, z^{\left ( a^{2}+ab+b^{2} \right )}=k\times 0$

বা, $log\, x^{\left ( b^{2}+bc+c^{2} \right )}.\, y^{\left ( c^{2}+ca+a^{2} \right )}.\, z^{\left ( a^{2}+ab+b^{2} \right )}=0$

বা, $log\, x^{\left ( b^{2}+bc+c^{2} \right )}.\, y^{\left ( c^{2}+ca+a^{2} \right )}.\, z^{\left ( a^{2}+ab+b^{2} \right )}=log\, 1$

${\color{DarkGreen} \therefore x^{b^{2}+bc+c^{2}}.\, y^{c^{2}+ca+a^{2}}.\, z^{a^{2}+ab+b^{2}}=1}$ (প্রমাণিত)

9. যদি a^{3 − x}. b^5x= a^{5 + x}. b^3x হয়, তাহলে দেখাই যে ${\color{Blue} xlog\left ( \frac{b}{a} \right )=loga}$

সমাধানঃ

a^{3 − x}. b^5x= a^{5 + x}. b^3x

বা, $\frac{b^{5x}}{b^{3x}}=\frac{a^{5+x}}{a^{3-x}}$

বা, b^{5x − 3x}= a^{5+x − 3+x}

বা, b²^x= a^{2 + 2x}

বা, b²^x= a². a^2x

বা, $\frac{b^{2x}}{a^{2x}}=a^{2}$

বা, $\left (\frac{b}{a} \right )^{2x}=a^{2}$

a^{3 − x}. b^5x= a^{5 + x}. b^3x

বা, $\frac{b^{5x}}{b^{3x}}=\frac{a^{5+x}}{a^{3-x}}$

বা, b^{5x − 3x}= a^{5+x − 3+x} ${\color{Blue} \left [ \because log\, _{a}\, \frac{m}{n}=log\, _{a}\, m-log\, _{a}\, n \right ]}$

বা, b²^x= a^{2 + 2x}

বা, b²^x= a². a^2x ${\color{Blue} \left [ \because log\, _{a}\, mn=log\, _{a}\, m+log\, _{a}\, n \right ]}$

বা, $\frac{b^{2x}}{a^{2x}}=a^{2}$

বা, $\left (\frac{b}{a} \right )^{2x}=a^{2}$

বা, $\left [\left (\frac{b}{a} \right )^{x} \right ]^{2}=a^{2}$

বা, $\left (\frac{b}{a} \right )^{x}=a$

উভয়দিকে log নিয়ে পাই,

বা, $log\left (\frac{b}{a} \right )^{x}=log\, a$

${\color{DarkGreen} \therefore x\, log\left (\frac{b}{a} \right )=log\, a}$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$ (প্রমাণিত)

10. সমাধান করি :

(a) ${\color{Blue} log_{8}\left [ log_{2}\left \{ log_{3}\left ( 4^{x}+17 \right ) \right \} \right ]=\frac{1}{3}}$

সমাধানঃ

$log_{8}\left [ log_{2}\left \{ log_{3}\left ( 4^{x}+17 \right ) \right \} \right ]=\frac{1}{3}$

বা, $\left [ log_{2}\left \{ log_{3}\left ( 4^{x}+17 \right ) \right \} \right ]=8^{\frac{1}{3}}$ [যেহেতু, log_a b = c হলে, a^c = b হয়]

বা, $\left [ log_{2}\left \{ log_{3}\left ( 4^{x}+17 \right ) \right \} \right ]=\left (2^{3} \right )^{\frac{1}{3}}$

বা, $\left [ log_{2}\left \{ log_{3}\left ( 4^{x}+17 \right ) \right \} \right ]=2$

বা, $\left \{ log_{3}\left ( 4^{x}+17 \right ) \right \}=2^{2}$ [যেহেতু, log_a b = c হলে, a^c = b হয়]

বা, $\left \{ log_{3}\left ( 4^{x}+17 \right ) \right \}=4$

বা, $\left ( 4^{x}+17 \right )=3^{4}$ [যেহেতু, log_a b = c হলে, a^c = b হয়]

বা, 4^x + 17 = 81

বা, 4^x = 81 − 17

বা, 4^x = 64

বা, 4^x = 4³

∴ x = 3 (উত্তর)

10. সমাধান করি :

(b) log₈x + log₄x + log₂x = 11

সমাধানঃ

log₈x + log₄x + log₂x = 11

বা, $\frac{log\, x}{log\, 8}+\frac{log\, x}{log\, 4}+\frac{log\, x}{log\, 2}=11$

বা, $\frac{log\, x}{log\, 2^{3}}+\frac{log\, x}{log\, 2^{2}}+\frac{log\, x}{log\, 2}=11$

বা, $\frac{log\, x}{3log\, 2}+\frac{log\, x}{2log\, 2}+\frac{log\, x}{log\, 2}=11$

বা, $\frac{log\, x}{log\, 2}\times \left ( \frac{1}{3}+\frac{1}{2}+1 \right )=11$

বা, $\frac{log\, x}{log\, 2}\times \left ( \frac{2+3+6}{6} \right )=11$

বা, $\frac{log\, x}{log\, 2}\times \left ( \frac{11}{6} \right )=11$

বা, $\frac{log\, x}{log\, 2}=11\times \frac{6}{11}$

বা, $\frac{log\, x}{log\, 2}=6$

বা, log x = 6 log 2

বা, log x = log 2⁶ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

বা, x = 2⁶

∴ x = 64 (উত্তর)

11. দেখাই log₁₀ 2 -এর মান ${\color{Blue} \frac{1}{4}}$ এবং ${\color{Blue} \frac{1}{3}}$ -এর মধ্যে অবস্থিত।

সমাধানঃ

ধরি, x = log₁₀ 2

∴ 10^x = 2 [যেহেতু, log_a b = c হলে, a^c = b হয়]

বা, 10¹²^x = 2¹²

বা, 10¹²^x = 4096

এখন,

1000 < 4096 < 10000

= 10³< 4096 < 10⁴

= 10³< 10¹²^x< 10⁴

= 3 < 12x < 4

$=\frac{3}{12}<\frac{12x}{12}<\frac{4}{12}$

$=\frac{1}{4}<x<\frac{1}{3}$

$=\frac{1}{4}<log\, _{10}\, 2<\frac{1}{3}$ (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

(d) 16

সমাধানঃ

log_√x 0.25 = 4

বা, (√x)⁴ = 0.25 [যেহেতু, log_a b = c হলে, a^c = 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

(d) 18

সমাধানঃ

log₁₀ (7x − 5) = 2

বা, 10² = (7x − 5) [যেহেতু, log_a b = c হলে, a^c = 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) ${\color{Blue} \frac{1}{a}}$

(d) a

সমাধানঃ

log₈ 27

$=\frac{log\, 27}{log\, 8}$

$=\frac{log\, 3^{3}}{log\, 2^{3}}$

$=\frac{3log\, 3}{3log\, 2}$

$=\frac{log\, 3}{log\, 2}$

= log₂3

= a

উত্তরঃ (d) a

Koshe dekhi 21 class 9

12. বহু বিকল্পীয় প্রশ্ন (M.C.Q)

(iv) log_√2 x = a হলে, log_2√2 x হবে

(a) ${\color{Blue} \frac{a}{3}}$

(b) a

(d) 3a

সমাধানঃ

log_2√2 x

$=\frac{log\, x}{log\, 2\sqrt{2}}$

$=\frac{log\, x}{log\, \left (\sqrt{2} \right )^{3}}$

$=\frac{log\, x}{3log\, \sqrt{2}}$

$=\frac{1}{3}\times log\, _{\sqrt{2}}\, x$

$=\frac{1}{3}\times a$

$=\frac{a}{3}$

উত্তরঃ (a) ${\color{DarkGreen} \frac{a}{3}}$

Koshe dekhi 21 class 9

12. বহু বিকল্পীয় প্রশ্ন (M.C.Q)

(v) ${\color{Blue} log_{x}\frac{1}{3}=-\frac{1}{3}}$ হলে, x -এর মান হবে

(a) 27

(b) 9

(d) ${\color{Blue} \frac{1}{27}}$

সমাধানঃ

$log_{x}\, \frac{1}{3}=-\frac{1}{3}$

বা, ${x}^{-\frac{1}{3}}=\frac{1}{3}$ [যেহেতু, log_a b = c হলে, a^c = b হয়]

বা, ${x}^{-\frac{1}{3}}=3^{-1}$

বা, ${x}^{-\frac{1}{3}}=\left (3^{3} \right )^{-\frac{1}{3}}$

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

$=log_{4}\, log_{4}\, log_{4}\, 4^{4}$

$=log_{4}\, log_{4}\, 4log_{4}\, 4$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

$=log_{4}\, log_{4}\, 4\times 1$ ${\color{Blue} [\because log_{a}\, a=1]}$

$=log_{4}\, log_{4}\, 4$

$=log_{4}\, 1$ ${\color{Blue} [\because log_{a}\, a=1]}$

= 0 (উত্তর)

Koshe dekhi 21 class 9

13. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্ন :

(ii) ${\color{Blue} log\frac{a^{n}}{b^{n}}+log\frac{b^{n}}{c^{n}}+log\frac{c^{n}}{a^{n}}}$ -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 ${\color{Blue} a^{log_{a}x}=x}$

সমাধানঃ

ধরি, $k=a^{log_{a}x}$

উভয়পক্ষে log নিয়ে পাই,

$log\, k=log\, a^{log_{a}x}$

বা, $log\, k=log\, a\times log_{a}x$

বা, $log\, k=log\, a\times \frac{log\, x}{log\, a}$

বা, log k = log x

বা, k = x

${\color{DarkGreen} \therefore a^{log_{a}x}=x}$ (প্রমাণিত)

Koshe dekhi 21 class 9

13. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্ন :

(iv) log_e 2. log_x 25 = log₁₀ 16. log_e10 হলে, x -এর মান নির্ণয় করি।

সমাধানঃ

log_e 2. log_x 25 = log₁₀ 16. log_e10

বা, $log_{e}\, 2.log_{x}\, 25=\frac{log\, 16}{log\, 10}\times \frac{log\, 10}{log\, e}$

বা, $log_{e}\, 2.log_{x}\, 25=log_{e}\, 16$

বা, $log_{e}\, 2.log_{x}\, 25=log_{e}\, 2^{4}$

বা, $log_{e}\, 2.log_{x}\, 25=4\, log_{e}\, 2$ ${\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}$

বা, $log_{x}\, 25=\frac{4\, log_{e}\, 2}{log_{e}\, 2}$

বা, $log_{x}\, 25=4$

বা, x⁴ = 25

বা, x⁴ = (√5)⁴

∴ x = √5 (উত্তর)

Koshe dekhi 21 class 9

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