Class 9 Chapter 2 সূচকের নিয়মাবলী (Laws of Indices)

নবম শ্রেণী – দ্বিতীয় অধ্যায় : সূচকের নিয়মাবলি সম্পূর্ণ সমাধান

কষে দেখি – 2

 

1. মান নির্ণয় করি –

(i) {{\left( \sqrt[5]{8} \right)}^{\frac{5}{2}}}\times {{\left( 16 \right)}^{-\frac{3}{2}}}

\[={{\left( {{8}^{\frac{1}{5}}} \right)}^{\frac{5}{2}}}\times {{\left( {{2}^{4}} \right)}^{-\frac{3}{2}}}\]

\[={{\left\{ {{\left( {{2}^{3}} \right)}^{\frac{1}{5}}} \right\}}^{\frac{5}{2}}}\times {{2}^{4\times \left( -\frac{3}{2} \right)}}\]

\[={{\left\{ {{2}^{\frac{3}{5}}} \right\}}^{\frac{5}{2}}}\times {{2}^{-6}}\]

\[={{2}^{\frac{3}{5}\times \frac{5}{2}}}\times {{2}^{-6}}\]

\[={{2}^{\frac{3}{2}}}\times {{2}^{-6}}\]

\[={{2}^{\frac{3}{2}-6}}\]

\[={{2}^{-\frac{9}{2}}}\]

(ii)   \,{{\left\{ {{\left( 125 \right)}^{-2}}\times {{\left( 16 \right)}^{-\frac{3}{2}}} \right\}}^{-\frac{1}{6}}} 

\[={{\left\{ {{\left( {{5}^{3}} \right)}^{-2}}\times {{\left( {{2}^{4}} \right)}^{-\frac{3}{2}}} \right\}}^{-\frac{1}{6}}}\]

\[={{\left\{ {{5}^{3\left( -2 \right)}}\times {{2}^{4\left( -\frac{3}{2} \right)}} \right\}}^{-\frac{1}{6}}}\]

\[={{\left\{ {{5}^{-6}}\times {{2}^{-6}} \right\}}^{-\frac{1}{6}}}\]

\[={{5}^{-6\left( -\frac{1}{6} \right)}}\times {{2}^{-6\left( -\frac{1}{6} \right)}}\]

\[=5\times 2=10\]

(iii) {{4}^{\frac{1}{3}}}\times \left[ {{2}^{\frac{1}{3}}}\times {{3}^{\frac{1}{2}}} \right]\div {{9}^{\frac{1}{4}}} 

\[={{\left( {{2}^{2}} \right)}^{\frac{1}{3}}}\times \left[ {{2}^{\frac{1}{3}}}\times {{3}^{\frac{1}{2}}} \right]\div {{\left( {{3}^{2}} \right)}^{\frac{1}{4}}}\]

\[={{2}^{\frac{2}{3}}}\times {{2}^{\frac{1}{3}}}\times {{3}^{\frac{1}{2}}}\times {{\left( {{3}^{2}} \right)}^{-\frac{1}{4}}}\]

\[={{2}^{\frac{2}{3}}}\times {{2}^{\frac{1}{3}}}\times {{3}^{\frac{1}{2}}}\times {{3}^{-\frac{1}{2}}}\]

\[={{2}^{\frac{2}{3}+\frac{1}{3}}}\times {{3}^{\frac{1}{2}-\frac{1}{2}}}\]

\[={{2}^{1}}\times {{3}^{0}}=2\times 1=2\]

 

2. সরল করি –

(i)   {{\left( 8{{a}^{3}}\div 27{{x}^{-3}} \right)}^{\frac{2}{3}}}\times {{\left( 64{{a}^{3}}\div 27{{x}^{-3}} \right)}^{-\frac{2}{3}}} 

\[={{\left( {{2}^{3}}{{a}^{3}}\div {{3}^{3}}{{x}^{-3}} \right)}^{\frac{2}{3}}}\times {{\left( {{2}^{6}}{{a}^{3}}\div {{3}^{3}}{{x}^{-3}} \right)}^{-\frac{2}{3}}}\]

\[={{\left( {{2}^{3}}{{a}^{3}}\times {{3}^{-3}}{{x}^{3}} \right)}^{\frac{2}{3}}}\times {{\left( {{2}^{6}}{{a}^{3}}\times {{3}^{-3}}{{x}^{3}} \right)}^{-\frac{2}{3}}}\]

\[={{2}^{3\times \frac{2}{3}}}\times {{a}^{3\times \frac{2}{3}}}\times {{3}^{-3\times \frac{2}{3}}}\times {{x}^{3\times \frac{2}{3}}}\times {{2}^{6\times \left( -\frac{2}{3} \right)}}\times {{a}^{3\times \left( -\frac{2}{3} \right)}}\times {{3}^{-3\times \left( -\frac{2}{3} \right)}}\times {{x}^{3\times \left( -\frac{2}{3} \right)}}\] 

\[={{2}^{2}}\times {{a}^{2}}\times {{3}^{-2}}\times {{x}^{2}}\times {{2}^{-4}}\times {{a}^{-2}}\times {{3}^{2}}\times {{x}^{-2}}\] 

\[={{2}^{2-4}}\times {{3}^{-2+2}}\times {{a}^{2-2}}\times {{x}^{2-2}}\] 

\[={{2}^{-2}}\times {{3}^{0}}\times {{a}^{0}}\times {{x}^{0}}\] 

\[=\frac{1}{{{2}^{2}}}\times 1\times 1\times 1=\frac{1}{4}\] 

(ii)  {{\left\{ {{\left( {{x}^{-5}} \right)}^{\frac{2}{3}}} \right\}}^{\frac{-3}{10}}}

\[={{\left\{ {{x}^{\left( -5 \right)\times \frac{2}{3}}} \right\}}^{-\frac{3}{10}}}\]

\[={{\left\{ {{x}^{-\frac{10}{3}}} \right\}}^{-\frac{3}{10}}}\] 

\[={{x}^{\left( -\frac{10}{3} \right)\times \left( -\frac{3}{10} \right)}}=x\]

(iii) \,{{\left[ {{\left\{ {{\left( {{2}^{-1}} \right)}^{-1}} \right\}}^{-1}} \right]}^{-1}} 

\[={{\left[ {{\left\{ {{2}^{\left( -1 \right)\times \left( -1 \right)}} \right\}}^{-1}} \right]}^{-1}}\] 

\[={{\left[ {{\left\{ {{2}^{1}} \right\}}^{-1}} \right]}^{-1}}\] 

\[={{\left[ {{2}^{1\times \left( -1 \right)}} \right]}^{-1}}\] 

\[={{2}^{\left( -1 \right)\times \left( -1 \right)}}=2\]

(iv) \,\sqrt[3]{{{a}^{-2}}}.b\times \sqrt[3]{{{b}^{-2}}}.c\times \sqrt[3]{{{c}^{-2}}}.a 

\[={{a}^{\left( -2 \right)\times \frac{1}{3}}}\times b\times {{b}^{\left( -2 \right)\times \frac{1}{3}}}\times c\times {{c}^{\left( -2 \right)\times \frac{1}{3}}}\times a\]

\[={{a}^{-\frac{2}{3}}}\times b\times {{b}^{-\frac{2}{3}}}\times c\times {{c}^{-\frac{2}{3}}}\times a\]

\[={{a}^{-\frac{2}{3}+1}}\times {{b}^{-\frac{2}{3}+1}}\times {{c}^{-\frac{2}{3}+1}}\]

\[={{a}^{\frac{1}{3}}}\times {{b}^{\frac{1}{3}}}\times {{c}^{\frac{1}{3}}}\]

\[={{\left( abc \right)}^{\frac{1}{3}}}=\sqrt[3]{abc}\]

(v)  \,{{\left( \frac{{{4}^{m+\frac{1}{4}}}\times \sqrt{{{2.2}^{m}}}}{2.\sqrt{{{2}^{-m}}}} \right)}^{\frac{1}{m}}} 

\[={{\left( \frac{{{2}^{2\left( m+\frac{1}{4} \right)}}\times \sqrt{{{2}^{m+1}}}}{{{2.2}^{-\frac{m}{2}}}} \right)}^{\frac{1}{m}}}\]

\[={{\left( \frac{{{2}^{2m+\frac{1}{2}}}\times {{2}^{\frac{m+1}{2}}}}{{{2}^{1-\frac{m}{2}}}} \right)}^{\frac{1}{m}}}\]

\[={{\left( {{2}^{2m+\frac{1}{2}+\frac{m+1}{2}-1+\frac{m}{2}}} \right)}^{\frac{1}{m}}}\]

\[={{\left( {{2}^{\frac{4m+1+m+1-2+m}{2}}} \right)}^{\frac{1}{m}}}\]

\[={{\left( {{2}^{\frac{6m}{2}}} \right)}^{\frac{1}{m}}}\]

\[={{2}^{3m\times \frac{1}{m}}}={{2}^{3}}=8\]

(vi) {{9}^{-3}}\times \frac{{{16}^{\frac{1}{4}}}}{{{6}^{-2}}}\times {{\left( \frac{1}{27} \right)}^{-\frac{4}{3}}} 

\[={{\left( {{3}^{2}} \right)}^{-3}}\times \frac{{{\left( {{2}^{4}} \right)}^{\frac{1}{4}}}}{{{\left( 2\times 3 \right)}^{-2}}}\times {{\left( \frac{1}{{{3}^{3}}} \right)}^{-\frac{4}{3}}}\]

\[={{3}^{-6}}\times \frac{2}{{{2}^{-2}}\times {{3}^{-2}}}\times {{\left( {{3}^{-3}} \right)}^{-\frac{4}{3}}}\]

\[={{3}^{-6}}\times 2\times {{2}^{2}}\times {{3}^{2}}\times {{3}^{4}}\]

\[={{2}^{1+2}}\times {{3}^{-6+2+4}}\]

\[={{2}^{3}}\times {{3}^{0}}=8\times 1=8\]

(vii) {{\left( \frac{{{x}^{a}}}{{{x}^{b}}} \right)}^{{{a}^{2}}+ab+{{b}^{2}}}}\times {{\left( \frac{{{x}^{b}}}{{{x}^{c}}} \right)}^{{{b}^{2}}+bc+{{c}^{2}}}}\times {{\left( \frac{{{x}^{c}}}{{{x}^{a}}} \right)}^{{{c}^{2}}+ca+{{a}^{2}}}}

\[={{\left( {{x}^{a-b}} \right)}^{{{a}^{2}}+ab+{{b}^{2}}}}\times {{\left( {{x}^{b-c}} \right)}^{{{b}^{2}}+bc+{{c}^{2}}}}\times {{\left( {{x}^{c-a}} \right)}^{{{c}^{2}}+ca+{{a}^{2}}}}\]

\[={{x}^{\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)}}\times {{x}^{\left( b-c \right)\left( {{b}^{2}}+bc+{{c}^{2}} \right)}}\times {{x}^{\left( c-a \right)\left( {{c}^{2}}+ca+{{a}^{2}} \right)}}\]

\[={{x}^{{{a}^{3}}-{{b}^{3}}}}\times {{x}^{{{b}^{3}}-{{c}^{3}}}}\times {{x}^{{{c}^{3}}-{{a}^{3}}}}\]

\[={{x}^{{{a}^{3}}-{{b}^{3}}+{{b}^{3}}-{{c}^{3}}+{{c}^{3}}-{{a}^{3}}}}\]

\[={{x}^{0}}=1\]

 

3. মানের ঊর্ধবক্রমানুসারে সাজাই –

(i) {{5}^{\frac{1}{2}}},\,{{10}^{\frac{1}{4}}},\,{{6}^{\frac{1}{3}}}

উত্তর –  2, 4, 3 এর লসাগু = 12

\[{{5}^{\frac{1}{2}}}={{5}^{\frac{6}{12}}}={{\left( {{5}^{6}} \right)}^{\frac{1}{12}}}={{\left( 15625 \right)}^{\frac{1}{12}}}\]

\[{{10}^{\frac{1}{4}}}={{10}^{\frac{3}{12}}}={{\left( {{10}^{3}} \right)}^{\frac{1}{12}}}={{\left( 1000 \right)}^{\frac{1}{12}}}\]

\[{{6}^{\frac{1}{3}}}={{6}^{\frac{4}{12}}}={{\left( {{6}^{4}} \right)}^{\frac{1}{12}}}={{\left( 1296 \right)}^{\frac{1}{12}}}\]

\[{{\left( 15625 \right)}^{\frac{1}{12}}}>{{\left( 1296 \right)}^{\frac{1}{12}}}>{{\left( 1000 \right)}^{\frac{1}{12}}}\]

{{5}^{\frac{1}{2}}},\,{{10}^{\frac{1}{4}}},\,{{6}^{\frac{1}{3}}}কে মানের ঊর্ধবক্রমানুসারে সাজালে পাই – \,{{10}^{\frac{1}{4}}},\,{{6}^{\frac{1}{3}}},{{5}^{\frac{1}{2}}}

(ii) {{3}^{\frac{1}{3}}},\,{{2}^{\frac{1}{2}}},\,{{8}^{\frac{1}{4}}}

উত্তর –  3, 2, 4 এর লসাগু = 12

\[{{3}^{\frac{1}{3}}}={{3}^{\frac{4}{12}}}={{\left( {{3}^{4}} \right)}^{\frac{1}{12}}}={{\left( 81 \right)}^{\frac{1}{12}}}\]

\[{{2}^{\frac{1}{2}}}={{2}^{\frac{6}{12}}}={{\left( {{2}^{6}} \right)}^{\frac{1}{12}}}={{\left( 64 \right)}^{\frac{1}{12}}}\]

\[{{8}^{\frac{1}{4}}}={{8}^{\frac{3}{12}}}={{\left( {{8}^{3}} \right)}^{\frac{1}{12}}}={{\left( 512 \right)}^{\frac{1}{12}}}\]

\[{{\left( 512 \right)}^{\frac{1}{12}}}>{{\left( 81 \right)}^{\frac{1}{12}}}>{{\left( 64 \right)}^{\frac{1}{12}}}\]

{{3}^{\frac{1}{3}}},\,{{2}^{\frac{1}{2}}},\,{{8}^{\frac{1}{4}}} কে মানের ঊর্ধবক্রমানুসারে সাজালে পাই – {{2}^{\frac{1}{2}}},{{3}^{\frac{1}{3}}},\,{{8}^{\frac{1}{4}}}

(iii) {{2}^{60}},\,{{3}^{48}},\,{{4}^{36}},\,{{5}^{24}}

উত্তর –  60, 48, 36, 24 এর গসাগু = 6

\[{{2}^{60}}={{\left( {{2}^{10}} \right)}^{3}}={{\left( 1024 \right)}^{6}}\]

\[{{3}^{48}}={{\left( {{3}^{8}} \right)}^{6}}={{\left( 6561 \right)}^{6}}\]

\[{{4}^{36}}={{\left( {{4}^{6}} \right)}^{6}}={{\left( 4096 \right)}^{6}}\]

\[{{5}^{24}}={{\left( {{5}^{4}} \right)}^{6}}={{\left( 625 \right)}^{6}}\]

\[{{\left( 6561 \right)}^{6}}>{{\left( 4096 \right)}^{6}}>{{\left( 1024 \right)}^{6}}>{{\left( 625 \right)}^{6}}\]

{{2}^{60}},\,{{3}^{48}},\,{{4}^{36}},\,{{5}^{24}} কে মানের ঊর্ধবক্রমানুসারে সাজালে পাই – {{5}^{24}},{{2}^{60}},\,{{4}^{36}},{{3}^{48}}

 

4. প্রমাণ করি –

(i) \,{{\left( \frac{{{a}^{q}}}{{{a}^{r}}} \right)}^{p}}\times {{\left( \frac{{{a}^{r}}}{{{a}^{p}}} \right)}^{q}}\times {{\left( \frac{{{a}^{p}}}{{{a}^{q}}} \right)}^{r}}=1

উত্তর –

বামপক্ষ,

\[\,{{\left( \frac{{{a}^{q}}}{{{a}^{r}}} \right)}^{p}}\times {{\left( \frac{{{a}^{r}}}{{{a}^{p}}} \right)}^{q}}\times {{\left( \frac{{{a}^{p}}}{{{a}^{q}}} \right)}^{r}}\]

\[={{\left( {{a}^{q-r}} \right)}^{p}}\times {{\left( {{a}^{r-p}} \right)}^{q}}\times {{\left( {{a}^{p-q}} \right)}^{r}}\]

\[={{a}^{pq-pr}}\times {{a}^{qr-pq}}\times {{a}^{pr-qr}}\]

\[={{a}^{pq-pr+qr-pq+pr-qr}}\]

\[={{a}^{0}}=1\]

বামপক্ষ = ডানপক্ষ (প্রমানিত)।

(ii) \,{{\left( \frac{{{x}^{m}}}{{{x}^{n}}} \right)}^{m+n}}{{\left( \frac{{{x}^{n}}}{{{x}^{l}}} \right)}^{n+l}}{{\left( \frac{{{x}^{l}}}{{{x}^{m}}} \right)}^{l+m}}

উত্তর –

বামপক্ষ,

\[{{\left( \frac{{{x}^{m}}}{{{x}^{n}}} \right)}^{m+n}}{{\left( \frac{{{x}^{n}}}{{{x}^{l}}} \right)}^{n+l}}{{\left( \frac{{{x}^{l}}}{{{x}^{m}}} \right)}^{l+m}}\]

\[={{\left( {{x}^{m-m}} \right)}^{m+n}}\times {{\left( {{x}^{n-l}} \right)}^{n+l}}\times {{\left( {{x}^{l-m}} \right)}^{l+m}}\]

\[={{x}^{\left( m-n \right)\left( m+n \right)}}\times {{x}^{\left( n-l \right)\left( n+l \right)}}\times {{x}^{\left( l-m \right)\left( l+m \right)}}\]

\[={{x}^{{{m}^{2}}-{{n}^{2}}}}\times {{x}^{{{n}^{2}}-{{l}^{2}}}}\times {{x}^{{{l}^{2}}-{{m}^{2}}}}\]

\[={{x}^{{{m}^{2}}-{{n}^{2}}+{{n}^{2}}-{{l}^{2}}+{{l}^{2}}-{{m}^{2}}}}\]

\[={{x}^{0}}=1\]

বামপক্ষ = ডানপক্ষ (প্রমানিত)।

(iii) {{\left( \frac{{{x}^{m}}}{{{x}^{n}}} \right)}^{m+n-l}}\times {{\left( \frac{{{x}^{n}}}{{{x}^{l}}} \right)}^{n+l-m}}\times {{\left( \frac{{{x}^{l}}}{{{x}^{m}}} \right)}^{l+m-n}}=1

উত্তর –

বামপক্ষ,

\[{{\left( \frac{{{x}^{m}}}{{{x}^{n}}} \right)}^{m+n-l}}\times {{\left( \frac{{{x}^{n}}}{{{x}^{l}}} \right)}^{n+l-m}}\times {{\left( \frac{{{x}^{l}}}{{{x}^{m}}} \right)}^{l+m-n}}\]

\[={{\left( {{x}^{m-n}} \right)}^{m+n-l}}\times {{\left( {{x}^{n-l}} \right)}^{n+l-m}}\times {{\left( {{x}^{l-m}} \right)}^{l+m-n}}\]

\[={{x}^{\left( m-n \right)\left( m+n-l \right)}}\times {{x}^{\left( n-l \right)\left( n+l-m \right)}}\times {{x}^{\left( l-m \right)\left( l+m-n \right)}}\]

\[={{x}^{\left( m-n \right)\left( m+n \right)-l\left( m-n \right)+\left( n-l \right)\left( n+l \right)-m\left( n-l \right)+\left( l-m \right)\left( l+m \right)-n\left( l-m \right)}}\]

\[={{x}^{{{m}^{2}}-{{n}^{2}}-lm+nl+{{n}^{2}}-{{l}^{2}}-mn+lm+{{l}^{2}}-{{m}^{2}}-nl+mn}}\]

\[={{x}^{0}}=1\]

বামপক্ষ = ডানপক্ষ (প্রমানিত)।

(iv) \,{{\left( {{a}^{\frac{1}{x-y}}} \right)}^{\frac{1}{x-z}}}\times {{\left( {{a}^{\frac{1}{y-z}}} \right)}^{\frac{1}{y-x}}}\times {{\left( {{a}^{\frac{1}{z-x}}} \right)}^{\frac{1}{z-y}}}=1

উত্তর –

বামপক্ষ,

\[\,{{\left( {{a}^{\frac{1}{x-y}}} \right)}^{\frac{1}{x-z}}}\times {{\left( {{a}^{\frac{1}{y-z}}} \right)}^{\frac{1}{y-x}}}\times {{\left( {{a}^{\frac{1}{z-x}}} \right)}^{\frac{1}{z-y}}}\]

\[={{a}^{\frac{1}{\left( x-y \right)\left( x-z \right)}}}\times {{a}^{\frac{1}{\left( y-z \right)\left( y-x \right)}}}\times {{a}^{\frac{1}{\left( z-x \right)\left( z-y \right)}}}\]

\[={{a}^{\frac{1}{\left( x-y \right)\left( x-z \right)}+\frac{1}{\left( y-z \right)\left( y-x \right)}+\frac{1}{\left( z-x \right)\left( z-y \right)}}}\]

\[={{a}^{\frac{-1}{\left( x-y \right)\left( z-x \right)}-\frac{1}{\left( y-z \right)\left( x-y \right)}-\frac{1}{\left( z-x \right)\left( y-z \right)}}}\]

\[={{a}^{\frac{-y+z-z+x-x+y}{\left( x-y \right)\left( y-z \right)\left( z-x \right)}}}\]

\[={{a}^{0}}=1\]

বামপক্ষ = ডানপক্ষ (প্রমানিত)।

 

5. x + z = 2y এবং b2 = ac হলে, দেখাই যে, {{a}^{y-z}}{{b}^{z-x}}{{c}^{x-y}}=1

উত্তর –

\[{{b}^{2}}=ac\]

\[\Rightarrow b={{\left( ac \right)}^{\frac{1}{2}}}={{a}^{\frac{1}{2}}}{{c}^{\frac{1}{2}}}\]

বামপক্ষ,

\[{{a}^{y-z}}{{b}^{z-x}}{{c}^{x-y}}\]

\[={{a}^{y-z}}{{\left( {{a}^{\frac{1}{2}}}{{c}^{\frac{1}{2}}} \right)}^{z-x}}{{c}^{x-y}}\,\,\left[ \because \,b={{a}^{\frac{1}{2}}}{{c}^{\frac{1}{2}}} \right]\]

\[={{a}^{y-z}}.{{a}^{\frac{z-x}{2}}}.{{c}^{\frac{z-x}{2}}}.{{c}^{x-y}}\]

\[={{a}^{\left( y-z \right)+\frac{z-x}{2}}}{{c}^{\left( x-y \right)+\frac{z-x}{2}}}\]

\[={{a}^{\frac{2y-2z+z-x}{2}}}{{c}^{\frac{2x-2y+z-x}{2}}}\]

\[={{a}^{\frac{2y-z-x}{2}}}{{c}^{\frac{x+z-2y}{2}}}\]

\[={{a}^{\frac{x+z-z-x}{2}}}{{c}^{\frac{x+z-x-z}{2}}}\]

\[={{a}^{0}}{{c}^{0}}=1\times 1=1\]

বামপক্ষ = ডানপক্ষ (প্রমানিত)।

 

6.  a=x{{y}^{p-1}},\,b=x{{y}^{q-1}} এবং c=x{{y}^{r-1}}  হলে, দেখাই যে, {{a}^{q-r}}{{b}^{r-p}}{{c}^{p-q}}=1 

উত্তর –

বামপক্ষ,

\[{{a}^{q-r}}{{b}^{r-p}}{{c}^{p-q}}\]

\[={{\left( x{{y}^{p-1}} \right)}^{q-r}}{{\left( x{{y}^{q-1}} \right)}^{r-p}}{{\left( x{{y}^{r-1}} \right)}^{p-q}}\]

\[={{x}^{q-r}}{{y}^{\left( p-1 \right)\left( q-r \right)}}{{x}^{r-p}}{{y}^{\left( q-1 \right)\left( r-p \right)}}{{x}^{p-q}}{{y}^{\left( r-1 \right)\left( p-q \right)}}\]

\[={{x}^{q-r+r-p+p-q}}{{y}^{pq-pr-q+r+qr-pq-r+p+pr-qr-p+q}}\]

\[={{x}^{0}}{{y}^{0}}=1\]

বামপক্ষ = ডানপক্ষ (প্রমানিত)।

 

7.  {{x}^{\frac{1}{a}}}={{y}^{\frac{1}{b}}}={{z}^{\frac{1}{c}}}এবং xyz = 1 হলে, দেখাই যে,   a + b + c = 0

উত্তর –

\[{{x}^{\frac{1}{a}}}={{y}^{\frac{1}{b}}}={{z}^{\frac{1}{c}}}=k\]( ধরি, k≠0, 1, – 1)

\[\Rightarrow x={{k}^{a}};\,y={{k}^{b}};\,z={{k}^{c}}\]

দেওয়া আছে,

\[xyz=1\]

\[\Rightarrow {{k}^{a}}.{{k}^{b}}.{{k}^{c}}=1\]

\[\Rightarrow {{k}^{a+b+c}}={{k}^{0}}\]

\[\therefore \,\,a+b+c=0\] (প্রমানিত)

 

8. ax = by = cz এবং abc = 1 হলে, দেখাই যে,  xy + yz + zx = 0

উত্তর –

\[{{a}^{x}}={{b}^{y}}={{c}^{z}}=k\] ( ধরি, k≠0, 1, – 1)

\[\Rightarrow a={{k}^{\frac{1}{x}}},\,b={{k}^{\frac{1}{y}}},\,c={{k}^{\frac{1}{z}}}\]

দেওয়া আছে,

\[abc=1\]

\[\Rightarrow {{k}^{\frac{1}{x}}}.{{k}^{\frac{1}{y}}}.{{k}^{\frac{1}{z}}}=1\]

\[\Rightarrow {{k}^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}}=1\]

\[\Rightarrow {{k}^{\frac{yz+zx+xy}{xyz}}}={{k}^{0}}\]

\[\Rightarrow \frac{yz+zx+xy}{xyz}=0\]

\[\therefore \,xy+yz+zx=0\](প্রমানিত)

 

9. সমাধান করি –

(i) \,{{49}^{x}}={{7}^{3}}

\[\Rightarrow {{\left( {{7}^{2}} \right)}^{x}}={{7}^{3}}\]

\[\Rightarrow {{7}^{2x}}={{7}^{3}}\]

\[\Rightarrow 2x=3\]

\[\therefore \,\,x=\frac{3}{2}=1\frac{1}{2}\]

নির্ণেয় সমাধান x=1\frac{1}{2}

(ii) \,{{2}^{x+2}}+{{2}^{x-1}}=9

\[\Rightarrow {{2}^{x}}{{.2}^{2}}+{{2}^{x}}{{.2}^{-1}}=9\]

\[\Rightarrow {{4.2}^{x}}+\frac{{{2}^{x}}}{2}=9\]

\[\Rightarrow \frac{{{8.2}^{x}}+{{2}^{x}}}{2}=9\]

\[\Rightarrow {{9.2}^{x}}=18\]

\[\Rightarrow {{2}^{x}}=\frac{18}{9}\]

\[\Rightarrow {{2}^{x}}=2\]

\[\therefore \,\,x=1\]

নির্ণেয় সমাধান x=1

(iii) \,\,{{2}^{x+1}}+{{2}^{x+2}}=48

\[\Rightarrow {{2}^{x}}.2+{{2}^{x}}{{.2}^{2}}=48\]

\[\Rightarrow {{2.2}^{x}}+{{4.2}^{x}}=48\]

\[\Rightarrow {{6.2}^{x}}=48\]

\[\Rightarrow {{2}^{x}}=\frac{48}{6}\]

\[\Rightarrow {{2}^{x}}=8\]

\[\Rightarrow {{2}^{x}}={{2}^{3}}\]

\[\therefore \,\,x=3\]

নির্ণেয় সমাধান x=3

(iv) \,\,{{2}^{4x}}{{.4}^{3x-1}}=\frac{{{4}^{2x}}}{{{2}^{3x}}}

\[\Rightarrow {{2}^{4x}}.{{\left( {{2}^{2}} \right)}^{3x-1}}=\frac{{{\left( {{2}^{2}} \right)}^{2x}}}{{{2}^{3x}}}\]

\[\Rightarrow {{2}^{4x}}{{.2}^{6x-2}}=\frac{{{2}^{4x}}}{{{2}^{3x}}}\]

\[\Rightarrow {{2}^{4x+6x-2}}={{2}^{4x-3x}}\]

\[\Rightarrow {{2}^{10x-2}}={{2}^{x}}\]

\[\Rightarrow 10x-2=x\]

\[\Rightarrow 10x-x=2\]

\[\Rightarrow 9x=2\]

\[\therefore \,\,x=\frac{2}{9}\]

নির্ণেয় সমাধান x=\frac{2}{9}

(v) \,\,9\times {{81}^{x}}={{27}^{2-x}}

\[\Rightarrow {{3}^{2}}\times {{\left( {{3}^{4}} \right)}^{x}}={{\left( {{3}^{3}} \right)}^{2-x}}\]

\[\Rightarrow {{3}^{2}}\times {{3}^{4x}}={{3}^{6-3x}}\]

\[\Rightarrow {{3}^{2+4x}}={{3}^{6-3x}}\]

\[\Rightarrow 2+4x=6-3x\]

\[\Rightarrow 4x+3x=6-2\]

\[\Rightarrow 7x=4\]

\[\therefore x=\frac{4}{7}\]

নির্ণেয় সমাধান x=\frac{4}{7}

(vi) \,\,{{2}^{5x+4}}+{{2}^{9}}={{2}^{10}}

\[\Rightarrow {{2}^{5x}}{{.2}^{4}}+{{2}^{9}}={{2}^{10}}\]

\[\Rightarrow {{2}^{5x}}{{.2}^{4}}={{2}^{10}}-{{2}^{9}}\]

\[\Rightarrow {{2}^{5x+4}}={{2}^{9}}\left( 2-1 \right)\]

\[\Rightarrow {{2}^{5x+4}}={{2}^{9}}\]

\[\Rightarrow 5x+4=9\]

\[\Rightarrow 5x=9-4\]

\[\Rightarrow 5x=5\]

\[\therefore \,\,x=1\]

নির্ণেয় সমাধান x=1

(vii) \,\,{{6}^{2x+4}}={{3}^{3x}}{{.2}^{x+8}}

\[\Rightarrow {{\left( 2\times 3 \right)}^{2x+4}}={{3}^{3x}}{{.2}^{x+8}}\]

\[\Rightarrow {{2}^{2x+4}}{{.3}^{2x+4}}={{3}^{3x}}{{.2}^{x+8}}\]

\[\Rightarrow \frac{{{2}^{2x+4}}}{{{2}^{x+8}}}=\frac{{{3}^{3x}}}{{{3}^{2x+4}}}\]

\[\Rightarrow {{2}^{2x+4-x-8}}={{3}^{3x-2x-4}}\]

\[\Rightarrow {{2}^{x-4}}={{3}^{x-4}}\]

\[\Rightarrow \frac{{{2}^{x-4}}}{{{3}^{x-4}}}=1\]

\[\Rightarrow {{\left( \frac{2}{3} \right)}^{x-4}}={{\left( \frac{2}{3} \right)}^{0}}\]

\[\Rightarrow x-4=0\]

\[\therefore \,\,x=4\]

নির্ণেয় সমাধান x=4

 

10. বহু বিকল্পীয় প্রশ্ন

(i)  {{\left( 0.243 \right)}^{0.2}}\times {{\left( 10 \right)}^{0.6}} -এর মান

(a) 0.3   (b) 3   (c) 0.9   (d) 9 

উত্তর –

\[{{\left( 0.243 \right)}^{0.2}}\times {{\left( 10 \right)}^{0.6}}\]

\[={{\left( \frac{243}{1000} \right)}^{0.2}}\times {{\left( 10 \right)}^{0.6}}\]

\[={{\left( \frac{{{3}^{5}}}{{{10}^{3}}} \right)}^{0.2}}\times {{\left( 10 \right)}^{0.6}}\]

\[=\frac{{{3}^{5\times 0.2}}}{{{10}^{3\times 0.2}}}\times {{\left( 10 \right)}^{0.6}}\]

\[=\frac{{{3}^{1.0}}}{{{10}^{0.6}}}\times {{\left( 10 \right)}^{0.6}}=3\]

{{\left( 0.243 \right)}^{0.2}}\times {{\left( 10 \right)}^{0.6}}  -এর মান (b) 3  

(ii) {{2}^{\frac{1}{2}}}\times {{2}^{-\frac{1}{2}}}\times {{\left( 16 \right)}^{\frac{1}{2}}}-এর মান

(a) 1  (b) 2   (c) 4  (d)   

উত্তর –

\[{{2}^{\frac{1}{2}}}\times {{2}^{-\frac{1}{2}}}\times {{\left( 16 \right)}^{\frac{1}{2}}}\]

\[={{2}^{\frac{1}{2}}}\times {{2}^{-\frac{1}{2}}}\times {{\left( {{2}^{4}} \right)}^{\frac{1}{2}}}\]

\[={{2}^{\frac{1}{2}-\frac{1}{2}+2}}={{2}^{2}}=4\]

{{2}^{\frac{1}{2}}}\times {{2}^{-\frac{1}{2}}}\times {{\left( 16 \right)}^{\frac{1}{2}}} -এর মান (c) 4 

(iii)  {{4}^{x}}={{8}^{3}}হলে, x -এর মান

(a) \frac{3}{2}  (b) \frac{9}{2}   (c) 3  (d) 9 

উত্তর –

\[{{4}^{x}}={{8}^{3}}\]

\[\Rightarrow {{\left( {{2}^{2}} \right)}^{x}}={{\left( {{2}^{3}} \right)}^{3}}\]

\[\Rightarrow {{2}^{2x}}={{2}^{9}}\]

\[\Rightarrow 2x=9\]

\[\therefore \,\,x=\frac{9}{2}\]

{{4}^{x}}={{8}^{3}}হলে, x -এর মান (b) \frac{9}{2} 

(iv)  {{20}^{-x}}=\frac{1}{7}হলে {{\left( 20 \right)}^{2x}}[/katex -এর মান </p> <p>(a) [katex]\frac{1}{49}  (b) 7  (c) 49  (d) 1 

উত্তর –

\[{{\left( 20 \right)}^{2x}}={{\left( 20 \right)}^{\left( -x \right)\times \left( -2 \right)}}={{\left( {{20}^{-x}} \right)}^{-2}}={{\left( \frac{1}{7} \right)}^{-2}}={{7}^{2}}=49\]

{{\left( 20 \right)}^{2x}}-এর মান  (c) 49 

(v)  4\times {{5}^{x}}=500হলে,{{x}^{x}} -এর মান 

(a) 8  (b) 1  (c) 64  (d) 27 

উত্তর –

\[4\times {{5}^{x}}=500\]

\[\Rightarrow {{5}^{x}}=\frac{500}{4}=125\]

\[\Rightarrow {{5}^{x}}={{5}^{3}}\]

\[\therefore \,\,x=3\]

\[\therefore \,\,{{x}^{x}}={{3}^{3}}=27\]

{{x}^{x}} -এর মান  (d) 27 

 

11. সংক্ষিপ্ত উত্তর ভিত্তিক প্রশ্ন –

(i) {{\left( 27 \right)}^{x}}={{\left( 81 \right)}^{y}} হলে, x:y কত হয় লিখি।

উত্তর –

\[{{\left( 27 \right)}^{x}}={{\left( 81 \right)}^{y}}\]

\[\Rightarrow {{\left( {{3}^{3}} \right)}^{x}}={{\left( {{3}^{4}} \right)}^{y}}\]

\[\Rightarrow {{3}^{3x}}={{3}^{4y}}\] 

\[\Rightarrow 3x=4y\] 

\[\Rightarrow \frac{x}{y}=\frac{4}{3}\] 

\[\therefore \,\,x:y=4:3\]

(ii) {{\left( {{5}^{5}}+0.01 \right)}^{2}}-{{\left( {{5}^{5}}-0.01 \right)}^{2}}={{5}^{x}} হলে, x –এর মান কত হিসাব করে লিখি।

উত্তর –

আমরা জানি,

\[{{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}=4ab\]

\[\therefore {{\left( {{5}^{5}}+0.01 \right)}^{2}}-{{\left( {{5}^{5}}-0.01 \right)}^{2}}={{5}^{x}}\]

\[\Rightarrow 4\times {{5}^{5}}\times 0.01={{5}^{x}}\]

\[\Rightarrow 4\times {{5}^{5}}\times \frac{1}{100}={{5}^{x}}\]

\[\Rightarrow \frac{{{5}^{5}}}{25}={{5}^{x}}\]

\[\Rightarrow \frac{{{5}^{5}}}{{{5}^{2}}}={{5}^{x}}\]

\[\Rightarrow {{5}^{3}}={{5}^{x}}\]

\[\therefore \,\,x=3\]

(iii) 3\times {{27}^{x}}={{9}^{x+4}}হলে, x –এর মান কত হিসাব করে লিখি।

উত্তর –

\[3\times {{27}^{x}}={{9}^{x+4}}\]

\[\Rightarrow 3\times {{\left( {{3}^{3}} \right)}^{x}}={{\left( {{3}^{2}} \right)}^{x+4}}\]

\[\Rightarrow 3\times {{3}^{3x}}={{3}^{2x+8}}\]

\[\Rightarrow {{3}^{1+3x}}={{3}^{2x+8}}\]

\[\Rightarrow 1+3x=2x+8\]

\[\Rightarrow 3x-2x=8-1\]

\[\therefore x=7\]

(iv) \sqrt[3]{{{\left( \frac{1}{64} \right)}^{\frac{1}{2}}}} –এর মান কত হিসাব করে লিখি।

উত্তর –

\[\sqrt[3]{{{\left( \frac{1}{64} \right)}^{\frac{1}{2}}}}=\sqrt[3]{{{\left( \frac{1}{{{2}^{6}}} \right)}^{\frac{1}{2}}}}=\sqrt[3]{\frac{1}{{{2}^{6\times \frac{1}{2}}}}}={{\left( \frac{1}{{{2}^{3}}} \right)}^{\frac{1}{3}}}=\frac{1}{{{2}^{3\times \frac{1}{3}}}}=\frac{1}{2}\]

(v)  {{3}^{{{3}^{3}}}} এবং {{\left( {{3}^{3}} \right)}^{3}}-এর মধ্যে কোনটি বৃহত্তর যুক্তিসহ লিখি। 

উত্তর –

\[{{3}^{{{3}^{3}}}}={{3}^{27}};\,{{\left( {{3}^{3}} \right)}^{3}}={{3}^{9}}\] 

\[\because \,{{3}^{27}}>{{3}^{9}}\] 

\[\therefore \,{{3}^{{{3}^{3}}}}>\,{{\left( {{3}^{3}} \right)}^{3}}\]

{{3}^{{{3}^{3}}}} বৃহত্তর।

Leave a Comment

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

0

Scroll to Top