Class 9 Chapter 21 লগারিদম (Logarithm)

নবম শ্রেণী – অধ্যায় ২১ : লগারিদম সম্পূর্ণ সমাধান

কষে দেখি – 21

1. মান নির্ণয় করি –

 

(i) {{\log }_{4}}\left( \frac{1}{64} \right)       

উত্তর-

সুতরাং, {{\log }_{4}}\left( \frac{1}{64} \right)

= {{\log }_{4}}1-{{\log }_{4}}64\,\,\left[ \because {{\log }_{a}}\left( \frac{M}{N} \right)={{\log }_{a}}M-{{\log }_{a}}N \right]

= 0-{{\log }_{4}}{{4}^{3}}\,\,\left[ \because {{\log }_{a}}1=0 \right]

= -3{{\log }_{4}}4

= -3\times 1=-3

 

(ii) {{\log }_{0.01}}0.000001                

উত্তর-

সুতরাং, {{\log }_{0.01}}0.000001

= {{\log }_{0.01}}{{\left( 0.01 \right)}^{3}}

= 3{{\log }_{0.01}}0.01=3

 

(iii) {{\log }_{\sqrt{6}}}216                  

উত্তর-

সুতরাং, {{\log }_{\sqrt{6}}}216

= {{\log }_{\sqrt{6}}}{{\left( 6 \right)}^{3}}

= {{\log }_{\sqrt{6}}}{{\left\{ {{\left( \sqrt{6} \right)}^{2}} \right\}}^{3}}

= {{\log }_{\sqrt{6}}}{{\left( \sqrt{6} \right)}^{6}}

= 6{{\log }_{\sqrt{6}}}\sqrt{6}=6\times 1=6

 

(iv) {{\log }_{2\sqrt{3}}}1728

উত্তর-

সুতরাং, {{\log }_{2\sqrt{3}}}1728

= {{\log }_{2\sqrt{3}}}\left( {{2}^{6}}\times {{3}^{3}} \right)

= {{\log }_{2\sqrt{3}}}{{2}^{6}}\times {{\left\{ {{\left( \sqrt{3} \right)}^{2}} \right\}}^{3}}

= {{\log }_{2\sqrt{3}}}{{2}^{6}}\times {{\left( \sqrt{3} \right)}^{6}}

= {{\log }_{2\sqrt{3}}}{{\left( 2\sqrt{3} \right)}^{6}}

= 6{{\log }_{2\sqrt{3}}}\left( 2\sqrt{3} \right)=6\times 1=6

 

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

উত্তর-

মনেকরি, নিধান x

সুতরাং, {{\log }_{x}}625=4

বা, {{x}^{4}}=625

বা, {{x}^{4}}={{5}^{4}}

বা, x=5

সুতরাং, নির্ণেয় নিধান হল 5।

 

(b) 5832–এর লগারিদম 6 হলে, নিধান কী হবে হিসাব করে লিখি।

উত্তর-

মনেকরি, নিধান x

সুতরাং, {{\log }_{x}}5832=6

বা, {{x}^{6}}=5832

বা, {{x}^{6}}={{2}^{3}}\times {{3}^{6}}

বা, {{x}^{6}}={{\left\{ {{\left( \sqrt{2} \right)}^{2}} \right\}}^{3}}\times {{3}^{6}}

বা, {{x}^{6}}={{\left( \sqrt{2} \right)}^{6}}\times {{3}^{6}}

বা, {{x}^{6}}={{\left( 3\sqrt{2} \right)}^{6}}

বা, x=3\sqrt{2}

সুতরাং, নির্ণেয় নিধান হল 3\sqrt{2}

 

3. (a) 1+{{\log }_{10}}a=2{{\log }_{10}}bহলে, a কে b –এর দ্বারা প্রকাশ করি।

উত্তর-

দেওয়া আছে, 1+{{\log }_{10}}a=2{{\log }_{10}}b

বা, {{\log }_{10}}10+{{\log }_{10}}a={{\log }_{10}}{{b}^{2}}\,\,\left[ \because {{\log }_{10}}10=1 \right]

বা, {{\log }_{10}}\left( 10\times a \right)={{\log }_{10}}{{b}^{2}}\,\,\left[ \because {{\log }_{a}}M+{{\log }_{a}}N={{\log }_{a}}MN \right]

বা, 10a={{b}^{2}}

বা, a=\frac{{{b}^{2}}}{10}

 

(b) 3+{{\log }_{10}}x=2{{\log }_{10}}y হলে, x কে y –এর দ্বারা প্রকাশ করি।

উত্তর-

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

বা, 3{{\log }_{10}}10+{{\log }_{10}}x=2{{\log }_{10}}y\,\,\left[ \because {{\log }_{10}}10=1 \right]

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

বা, {{\log }_{10}}{{10}^{3}}\times x={{\log }_{10}}{{y}^{2}}

বা, 100x={{y}^{2}}

বা, x=\frac{{{y}^{2}}}{1000}

 

4. মান নির্ণয় করি –

(a) {{\log }_{2}}\left[ {{\log }_{2}}\left\{ {{\log }_{3}}\left( {{\log }_{3}}{{27}^{3}} \right) \right\} \right]

উত্তর-

সুতরাং, {{\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}}{{\left( {{3}^{3}} \right)}^{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]\,\,\left[ \because {{\log }_{a}}{{M}^{n}}=n{{\log }_{a}}M \right]

= {{\log }_{2}}\left[ {{\log }_{2}}\left\{ {{\log }_{3}}9 \right\} \right]\,\,\left[ \because {{\log }_{3}}3=1 \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]\,\,

= {{\log }_{2}}\left[ {{\log }_{2}}2 \right]\,\,

= {{\log }_{2}}1

= 0

 

(b) \frac{\log \sqrt{27}+\log 8-\log \sqrt{1000}}{\log 1.2}

উত্তর-

সুতরাং, \frac{\log \sqrt{27}+\log 8-\log \sqrt{1000}}{\log 1.2}

= \frac{\log {{\left( 27 \right)}^{\frac{1}{2}}}+\log 8-\log {{\left( 1000 \right)}^{\frac{1}{2}}}}{\log 1.2}

= \frac{\log {{\left( {{3}^{3}} \right)}^{\frac{1}{2}}}+\log {{2}^{3}}-\log {{\left( {{10}^{3}} \right)}^{\frac{1}{2}}}}{\log 1.2}

= \frac{\log {{3}^{\frac{3}{2}}}+3\log 2-\log {{10}^{\frac{3}{2}}}}{\log 1.2}

= \frac{\frac{3}{2}\log 3+3\log 2-\frac{3}{2}\log 10}{\log 1.2}

= \frac{\frac{3}{2}\left( \log 3+2\log 2-\log 10 \right)}{\log 1.2}

= \frac{\frac{3}{2}\left( \log 3+\log {{2}^{2}}-\log 10 \right)}{\log 1.2}

= \frac{\frac{3}{2}\left( \log 3+\log 4-\log 10 \right)}{\log 1.2}

= \frac{\frac{3}{2}\left( \log 3\times 4-\log 10 \right)}{\log 1.2}

= \frac{\frac{3}{2}\left( \log 12-\log 10 \right)}{\log 1.2}

= \frac{\frac{3}{2}\log \frac{12}{10}}{\log 1.2}

= \frac{\frac{3}{2}\log 1.2}{\log 1.2}=\frac{3}{2}

 

(c) {{\log }_{3}}4\times {{\log }_{4}}5\times {{\log }_{5}}6\times {{\log }_{6}}7\times {{\log }_{7}}3

উত্তর-

সুতরাং, {{\log }_{3}}4\times {{\log }_{4}}5\times {{\log }_{5}}6\times {{\log }_{6}}7\times {{\log }_{7}}3

= \frac{{{\log }_{10}}4}{{{\log }_{10}}3}\times \frac{{{\log }_{10}}5}{{{\log }_{10}}4}\times \frac{{{\log }_{10}}6}{{{\log }_{10}}5}\times \frac{{{\log }_{10}}7}{{{\log }_{10}}6}\times \frac{{{\log }_{10}}3}{{{\log }_{10}}7}

= 1

 

(d) {{\log }_{10}}\frac{384}{5}+{{\log }_{10}}\frac{81}{32}+3{{\log }_{10}}\frac{5}{3}+{{\log }_{10}}\frac{1}{9}

উত্তর-

সুতরাং, {{\log }_{10}}\frac{384}{5}+{{\log }_{10}}\frac{81}{32}+3{{\log }_{10}}\frac{5}{3}+{{\log }_{10}}\frac{1}{9}

= \left( {{\log }_{10}}384-{{\log }_{10}}5 \right)+\left( {{\log }_{10}}81-{{\log }_{10}}32 \right)+3\left( {{\log }_{10}}5-{{\log }_{10}}3 \right)+\left( {{\log }_{10}}1-{{\log }_{10}}9 \right)

= {{\log }_{10}}\left( 3\times {{2}^{7}} \right)-{{\log }_{10}}5+{{\log }_{10}}{{3}^{4}}-{{\log }_{10}}{{2}^{5}}+3{{\log }_{10}}5-3{{\log }_{10}}3+0-{{\log }_{10}}{{3}^{2}}

= {{\log }_{10}}3+7{{\log }_{10}}2-{{\log }_{10}}5+4{{\log }_{10}}3-5{{\log }_{10}}2+3{{\log }_{10}}5-3{{\log }_{10}}3-2{{\log }_{10}}3

= 5{{\log }_{10}}3-5{{\log }_{10}}3+7{{\log }_{10}}2-5{{\log }_{10}}2-{{\log }_{10}}5+3{{\log }_{10}}5

= 2{{\log }_{10}}2+2{{\log }_{10}}5

= 2\left( {{\log }_{10}}2+{{\log }_{10}}5 \right)

= 2{{\log }_{10}}\left( 2\times 5 \right)

= 2{{\log }_{10}}10

= 2\times 1=2

 

5. প্রমাণ করি –

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

উত্তর-

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

= \left( \log 75-\log 16 \right)-2\left( \log 5-\log 9 \right)+\left( \log 32-\log 243 \right)

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

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

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

= \log 2 = ডানপক্ষ [প্রমাণিত]

 

(ii) {{\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

উত্তর-

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

= {{\log }_{10}}15\times {{\log }_{15}}\left( 15\times 30 \right)+\frac{1}{2}{{\log }_{10}}16\times {{\log }_{4}}\left( 4\times 7 \right)-{{\log }_{10}}6\times {{\log }_{6}}\left( 3\times 6\times 7 \right)

= {{\log }_{10}}15\times {{\log }_{15}}450+{{\log }_{10}}{{16}^{\frac{1}{2}}}\times {{\log }_{4}}28-{{\log }_{10}}6\times {{\log }_{6}}126

= {{\log }_{10}}15\times {{\log }_{15}}450+{{\log }_{10}}4\times {{\log }_{4}}28-{{\log }_{10}}6\times {{\log }_{6}}126

= {{\log }_{10}}450+{{\log }_{10}}28-{{\log }_{10}}126\,\,\left[ \because {{\log }_{a}}M={{\log }_{b}}M\times {{\log }_{a}}b \right]

= {{\log }_{10}}\left( 450\times 28 \right)-{{\log }_{10}}126

= {{\log }_{10}}\left( \frac{450\times 28}{126} \right)

= {{\log }_{10}}100

= {{\log }_{10}}{{10}^{2}}

= 2{{\log }_{10}}10=2 = ডানপক্ষ [প্রমাণিত]

 

(iii) {{\log }_{2}}{{\log }_{2}}{{\log }_{4}}256+2{{\log }_{\sqrt{2}}}2=5

উত্তর-

বামপক্ষ = {{\log }_{2}}{{\log }_{2}}{{\log }_{4}}256+2{{\log }_{\sqrt{2}}}2

= {{\log }_{2}}{{\log }_{2}}{{\log }_{4}}{{4}^{4}}+2{{\log }_{\sqrt{2}}}{{\left( \sqrt{2} \right)}^{2}}

= {{\log }_{2}}{{\log }_{2}}4{{\log }_{4}}4+2\times 2{{\log }_{\sqrt{2}}}\sqrt{2}\,\,\left[ \because {{\log }_{a}}{{M}^{n}}=n{{\log }_{a}}M \right]

= {{\log }_{2}}{{\log }_{2}}4+4\,\,\left[ \because {{\log }_{a}}a=1 \right]

= {{\log }_{2}}{{\log }_{2}}{{2}^{2}}+4

= {{\log }_{2}}2{{\log }_{2}}2+4

= {{\log }_{2}}2+4

= 1+4=5 = ডানপক্ষ [প্রমাণিত]

 

(iv) {{\log }_{{{x}^{2}}}}x\times {{\log }_{{{y}^{2}}}}y\times {{\log }_{{{z}^{2}}}}z=\frac{1}{8}

উত্তর-

বামপক্ষ = {{\log }_{{{x}^{2}}}}x\times {{\log }_{{{y}^{2}}}}y\times {{\log }_{{{z}^{2}}}}z

=\frac{1}{{{\log }_{x}}{{x}^{2}}}\times \frac{1}{{{\log }_{y}}{{y}^{2}}}\times \frac{1}{{{\log }_{z}}{{z}^{2}}}\,\,\left[ \because {{\log }_{a}}b=\frac{1}{{{\log }_{b}}a} \right]

= \frac{1}{2{{\log }_{x}}x}\times \frac{1}{2{{\log }_{y}}y}\times \frac{1}{2{{\log }_{z}}z}

= \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\,\,\left[ \because {{\log }_{x}}x={{\log }_{y}}y={{\log }_{z}}z=1 \right]

= \frac{1}{8}= ডানপক্ষ [প্রমাণিত]

 

(v) {{\log }_{{{b}^{3}}}}a\times {{\log }_{{{c}^{3}}}}b\times {{\log }_{{{a}^{3}}}}c=\frac{1}{27}

উত্তর-

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}}}\,\,\left[ \because {{\log }_{a}}b=\frac{\log b}{\log a} \right]

= \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]

 

(vi) \frac{1}{{{\log }_{xy}}\left( xyz \right)}+\frac{1}{{{\log }_{yz}}\left( xyz \right)}+\frac{1}{{{\log }_{zx}}\left( xyz \right)}=2

উত্তর-

বামপক্ষ = \frac{1}{{{\log }_{xy}}\left( xyz \right)}+\frac{1}{{{\log }_{yz}}\left( xyz \right)}+\frac{1}{{{\log }_{zx}}\left( xyz \right)}

={{\log }_{xyz}}xy+{{\log }_{xyz}}yz+{{\log }_{xyz}}zx\,\,\left[ \because {{\log }_{a}}b=\frac{1}{{{\log }_{b}}a} \right]

= {{\log }_{xyz}}\left( xy\times yz\times zx \right)

= {{\log }_{xyz}}{{x}^{2}}{{y}^{2}}{{z}^{2}}

= {{\log }_{xyz}}{{\left( xyz \right)}^{2}}

= 2{{\log }_{xyz}}xyz

= 2\times 1=2= ডানপক্ষ [প্রমাণিত]

 

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

উত্তর-

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

= \log \left\{ \frac{{{a}^{2}}}{bc}\times \frac{{{b}^{2}}}{ca}\times \frac{{{c}^{2}}}{ab} \right\}

= \log \left\{ \frac{{{a}^{2}}{{b}^{2}}{{c}^{2}}}{{{a}^{2}}{{b}^{2}}{{c}^{2}}} \right\}

= \log 1=0 = ডানপক্ষ [প্রমাণিত]

 

(viii) \,{{x}^{\log y-\log z}}\times {{y}^{\log z-\log x}}\times {{z}^{\log x-\log y}}=1

উত্তর-

মনেকরি, \,P={{x}^{\log y-\log z}}\times {{y}^{\log z-\log x}}\times {{z}^{\log x-\log y}}

বা, \log P=\log \left[ {{x}^{\log y-\log z}}\times {{y}^{\log z-\log x}}\times {{z}^{\log x-\log y}} \right]

বা, \log P=\log {{x}^{\log y-\log z}}+\log {{y}^{\log z-\log x}}+\log {{z}^{\log x-\log y}}

বা, \log P=\left( \log y-\log z \right)\log x+\left( \log z-\log x \right)\log y+\left( \log x-\log y \right)\log z

বা, \log P=\log x\log y-\log x\log z+\log y\log z-\log x\log y+\log x\log z-\log y\log z

বা, \log P=0

বা, \log P=\log 1

বা, P=1

সুতরাং, {{x}^{\log y-\log z}}\times {{y}^{\log z-\log x}}\times {{z}^{\log x-\log y}}=1 [প্রমাণিত]

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