Thu. Jul 18th, 2024

Koshe dekhi 21 class 9

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 log0.01 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

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} 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{\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 :

(c) {\color{Blue} log_{3}\, 4\times log_{4}\, 5\times log_{5}\, 6\times log_{6}\, 7\times log_{7}\, 3}

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) 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} 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 = 14a2 b2  হয়, তাহলে দেখাই যে log (a+ b) = log a + log b + 2 log 2

সমাধানঃ

a4 + b4 = 14a2 b2 

বা, (a+ b)− 2a2 b2 = 14a2 b2 

বা, (a+ b)= 14a2 b2 + 2a2 b2 

বা, (a+ b)= 16a2 b2

বা, (a+ b)= (4ab)2

বা, a+ b= 4ab

উভয়পক্ষে log নিয়ে পাই,

log (a+ b2) = log (4ab)

বা, log (a+ b2) = log 4 + log a + log b

বা, log (a+ b2) = log 22 + log a + log b

বা, log (a+ b2) = 2 log 2 + log a + log b  {\color{Blue} [\because log\, _{a}\, m^{c}=c\, log\, _{a}\, m]}

∴ log (a+ b) = 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

 xb + c . yc + a . za + 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 − a+ 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} \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 (b+ bc + c2) গুণ করে পাই,

বা, (b+ bc + c2)log x = k (b − c) (b+ 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  a3 − x . b5x = a5 + x . b3x  হয়, তাহলে দেখাই যে  {\color{Blue} xlog\left ( \frac{b}{a} \right )=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, b2= a2 + 2x

বা, b2= a. a2x

বা, \frac{b^{2x}}{a^{2x}}=a^{2}

বা, \left (\frac{b}{a} \right )^{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} \left [ \because log\, _{a}\, \frac{m}{n}=log\, _{a}\, m-log\, _{a}\, n \right ]}

বা, b2= a2 + 2x

বা, b2= a. a2x  {\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, loga b = c হলে, ac = 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, loga b = c হলে, ac = b হয়]

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

বা, \left ( 4^{x}+17 \right )=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) logx + logx + logx = 11

সমাধানঃ

 logx + logx + logx = 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

∴ 10x = 2 [যেহেতু, loga b = c হলে, ac = b হয়]

বা, 1012x = 212

বা, 1012x = 4096

এখন,

1000 < 4096 < 10000

= 10< 4096 < 104

= 10< 1012x < 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 logx 0.25 = 4 হয়, তাহলে x -এর মান

 (a) 0.5

 (b) 0.25

 (c) 4

 (d) 16

সমাধানঃ

 logx 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} \frac{1}{a}}

 (c) 2a

 (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}

= log3

= 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} \frac{a}{3}}

 (b) a

 (c) 2a

 (d) 3a

সমাধানঃ

 log2√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

 (c) 3

 (d) {\color{Blue} \frac{1}{27}}

সমাধানঃ

 log_{x}\, \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 (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) loge 2. logx 25 = log10 16. log10 হলে, x -এর মান নির্ণয় করি। 

সমাধানঃ

 loge 2. logx 25 = log10 16. log10

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

Support Me

If you like my work then you can Support me by contributing a small amount which will help me a lot to grow my Website. It’s a request to all of you. You can donate me through phone pay / Paytm/ Gpay  on this number 7980608289 or by the link below :

Subscribe my Youtube channel : Science Duniya in Bangla

and    Learning Science

and visit Our website : learningscience.co.in 

গণিত প্রকাশ (দশম শ্রেণী) সম্পূর্ণ সমাধান

গণিত প্রকাশ (নবম শ্রেণী) সম্পূর্ণ সমাধান

গণিত প্রভা (ষষ্ঠ শ্রেণী) সম্পূর্ণ সমাধান

জীবন বিজ্ঞান  (দশম শ্রেণী) (Life Science)

Thank You

Leave a Reply

Your email address will not be published. Required fields are marked *

Insert math as
Block
Inline
Additional settings
Formula color
Text color
#333333
Type math using LaTeX
Preview
\({}\)
Nothing to preview
Insert
error: Content is protected !!