কষে দেখি 2 || 2. সূচকের নিয়মাবলী || WBBSE Class 9 Math Solution

2. সূচকের নিয়মাবলী | Exercise 2 all solution | Ganit Prakash Class IX math solution | WBBSE Class 9 Math Solution in Bengali |  

 

কষে দেখি 2

 

1. মান নির্নয় করোঃ
(i)${\left(\sqrt[5]{8}\right)}^{\frac{5}{2}}\times {(16)}^{\frac{-3}{2}}$
সমাধানঃ 

${\left(\sqrt[5]{8}\right)}^{\frac{5}{2}}\times {(16)}^{\frac{-3}{2}}$
=${\left(8^{\frac{1}{5}}\right)}^{\frac{5}{2}}\times {(2^4)}^{\frac{-3}{2}}$
=$8^{\frac{1}{5}\times \frac{5}{2}}\times 2^{4\times (-\frac{3}{2})}$
=${\left(2^3\right)}^{\frac{1}{2}}\times 2^{-6}$
=$2^{\frac{3}{2}}\times 2^{-6}$
=$2^{\frac{3}{2}-6}$
=$2^{\frac{3-12}{2}}$
=$2^{-\frac{9}{2}}$


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

${\{{\left(125\right)}^{-2}\times {\left(16\right)}^{\frac{-3}{2}}\}}^{\frac{-1}{6}}$
=${\{{\left(5^3\right)}^{-2}\times {\left(2^4\right)}^{\frac{-3}{2}}\}}^{\frac{-1}{6}}$
=${\{5^{-6}\times 2^{-6}\}}^{\frac{-1}{6}}$
=${\left\{{(5\times 2)}^{-6}\right\}}^{\frac{-1}{6}}$
=${\left(10\right)}^{-6\times \left(\frac{-1}{6}\right)}$
=10


(iii) $4^{\frac{1}{3}}\times [2^{\frac{1}{3}}\times 3^{\frac{1}{2}}]÷9^{\frac{1}{4}}$

সমাধানঃ 
$4^{\frac{1}{3}}\times [2^{\frac{1}{3}}\times 3^{\frac{1}{2}}]÷9^{\frac{1}{4}}$
=${\left(2^2\right)}^{\frac{1}{3}}\times 2^{\frac{1}{3}}\times 3^{\frac{1}{2}}÷{\left(3^2\right)}^{\frac{1}{4}}$
=$2^{\frac{2}{3}}\times 2^{\frac{1}{3}}\times 3^{\frac{1}{2}}÷3^{\frac{1}{2}}$
=$2^{\frac{2}{3}+\frac{1}{3}}\times 3^{\frac{1}{2}-\frac{1}{2}}$
=$2^{\frac{3}{3}}\times 3^0$
=2


2. সরল করোঃ
(i) ${(8a^3÷27x^{-3})}^{\frac{2}{3}}\times {(64a^3÷27x^{-3})}^{\frac{-2}{3}}$

সমাধানঃ 
${(8a^3÷27x^{-3})}^{\frac{2}{3}}\times {(64a^3÷27x^{-3})}^{\frac{-2}{3}}$
=${\{{\left(2a\right)}^3÷\frac{27}{x^3}\}}^{\frac{2}{3}}\times {\{{\left(4a\right)}^3÷\frac{27}{x^3}\}}^{\frac{-2}{3}}$
=${\{{\left(2a\right)}^3\times {\left(\frac{x}{3}\right)}^3\}}^{\frac{2}{3}}\times {\{{\left(4a\right)}^3\times {\left(\frac{x}{3}\right)}^3\}}^{\frac{-2}{3}}$
=${\left\{{(2a\times \frac{x}{3})}^3\right\}}^{\frac{2}{3}}\times {\left\{{(4a\times \frac{x}{3})}^3\right\}}^{\frac{-2}{3}}$
=${(2a\times \frac{x}{3})}^{3\times \frac{2}{3}}\times {(4a\times \frac{x}{3})}^{3\times \frac{-2}{3}}$
=${\left(\frac{2ax}{3}\right)}^2\times {\left(\frac{4ax}{3}\right)}^{-2}$
=${\left(\frac{2ax}{3}\right)}^2\times {\left(\frac{3}{4ax}\right)}^2$
=${(\frac{2ax}{3}\times \frac{3}{4ax})}^2$
=${\left(\frac{1}{2}\right)}^2=\frac{1}{4}$



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

সমাধানঃ 
${\left\{{\left(x^{-5}\right)}^{\frac{2}{3}}\right\}}^{\frac{-3}{10}}$
=${\left\{x^{-5\times \frac{2}{3}}\right\}}^{\frac{-3}{10}}$
=${\left(x^{\frac{-10}{3}}\right)}^{\frac{-3}{10}}$
=${\left(x\right)}^{\left(\frac{-10}{3}\right)\times \left(\frac{-3}{10}\right)}$
=$x^1$

= x

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

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

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

সমাধানঃ 

$\sqrt[3]{a^{-2}}.b\times \sqrt[3]{b^{-2}}.c\times \sqrt[3]{c^{-2}}.a$

=$a^{\frac{-2}{3}}.b\times b^{\frac{-2}{3}}.c\times c^{\frac{-2}{3}}.a$

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

=$a^{\frac{-2+3}{3}}\times b^{\frac{-2+3}{3}}\times c^{\frac{-2+3}{3}}$

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

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

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

সমাধানঃ 
${\left(\frac{4^{m+\frac{1}{4}}\times \sqrt{{2.2}^m}}{2.\sqrt{2^{-m}}}\right)}^{\frac{1}{m}}$
=${\left(\frac{2^{2(m+\frac{1}{4})}\times {\left({2.2}^m\right)}^{\frac{1}{2}}}{{2.2}^{\frac{-m}{2}}}\right)}^{\frac{1}{m}}$
=${\left(\frac{2^{2m+\frac{2}{4}}\times 2^{\frac{1}{2}}{.2}^{\frac{m}{2}}}{{2.2}^{\frac{-m}{2}}}\right)}^{\frac{1}{m}}$
=${\left(\frac{2^{2m+\frac{1}{2}+\frac{1}{2}+\frac{m}{2}}}{{2.2}^{\frac{-m}{2}}}\right)}^{\frac{1}{m}}$
=${\left(2^{2m+1+\frac{m}{2}-1+\frac{m}{2}}\right)}^{\frac{1}{m}}$
=${\left(2^{3m}\right)}^{\frac{1}{m}}$
=$2^3$

= 8


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

সমাধানঃ 
$9^{-3}\times \frac{{16}^{\frac{1}{4}}}{6^{-2}}\times {\left(\frac{1}{27}\right)}^{-\frac{4}{3}}$
=${\left(3\right)}^{2\times (-3)}\times \frac{2^{4\times \frac{1}{4}}}{{(2\times 3)}^{-2}}\times {\left(\frac{1}{3}\right)}^{3\times (-\frac{4}{3})}$
=$3^{-6}\times \frac{2}{2^{-2}\times 3^{-2}}\times {\left(\frac{1}{3}\right)}^{-4}$
=$\frac{1}{3^6}\times 2\times 2^2\times 3^2\times 3^4$
=$\frac{2^{1+2}\times 3^{2+4}}{3^6}$
=$2^3\times 3^{6-6}$
=$8\times 3^0$
=8

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

সমাধানঃ 

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

=$x^{(a-b)(a^2+ab+b^2)}\times x^{(b-c)(b^2+bc+c^2)}\times x^{(c-a)(c^2+ca+a^2)}$

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

=$x^{a^3-b^3+b^3-c^3+.{\mathrm{c}}^3-a^3}$

=$x^0$

= 1

3. মানের উর্ধ্বানুসারে সাজাইঃ
(i) $5^{\frac{1}{2}},{10}^{\frac{1}{4}},6^{\frac{1}{3}}$
সমাধানঃ 
$5^{\frac{1}{2}}=5^{\frac{6}{12}}={\left(5^6\right)}^{\frac{1}{12}}={\left(15625\right)}^{\frac{1}{12}}$
${10}^{\frac{1}{4}}={10}^{\frac{3}{12}}={\left({10}^3\right)}^{\frac{1}{12}}={\left(1000\right)}^{\frac{1}{12}}$
$6^{\frac{1}{3}}=6^{\frac{4}{12}}={\left(6^4\right)}^{\frac{1}{12}}={\left(1296\right)}^{\frac{1}{12}}$
যেহেতু, 1000<1296<15625
$\therefore {10}^{\frac{1}{4}}<6^{\frac{1}{3}}<5^{\frac{1}{2}}$
$\therefore$ মানের উর্ধ্বক্রমে সাজিয়ে পাই 

${10}^{\frac{1}{4}},6^{\frac{1}{3}},5^{\frac{1}{2}}$

(ii) $3^{\frac{1}{3}},2^{\frac{1}{2}},8^{\frac{1}{4}}$
সমাধানঃ 
$3^{\frac{1}{3}}=3^{\frac{4}{12}}={\left(3^4\right)}^{\frac{1}{12}}={\left(81\right)}^{\frac{1}{12}}$
$2^{\frac{1}{2}}=2^{\frac{6}{12}}={\left(2^6\right)}^{\frac{1}{12}}={\left(64\right)}^{\frac{1}{12}}$
$8^{\frac{1}{4}}=8^{\frac{3}{12}}={\left(8^3\right)}^{\frac{1}{12}}={\left(512\right)}^{\frac{1}{12}}$
যেহেতু, 64<81<512
$\therefore 2^{\frac{1}{2}}<3^{\frac{1}{3}}<8^{\frac{1}{4}}$
$\therefore$ মানের উর্ধ্বক্রমে সাজিয়ে পাই 

$2^{\frac{1}{2}},3^{\frac{1}{3}},8^{\frac{1}{4}}$


(iii) $2^{60},3^{48},4^{36},5^{24}$
সমাধানঃ 
$2^{60}=2^{5.12}=(2^5{)}^{12}=(32{)}^{12}$
$3^{48}=3^{4.12}=(3^4{)}^{12}=(81{)}^{12}$
$4^{36}=4^{3.12}=(4^3{)}^{12}=(64)12$
$5^{24}=5^{2.12}=(5^2{)}^{12}=(25{)}^{12}$
যেহেতু, 25<32<64<81
∴ $5^{24}<2^{60}<4^{36}<3^{48}$

∴  মানের উর্ধ্বক্রমে সাজিয়ে পাই $5^{24},2^{60},4^{36},3^{48}$
 
4. প্রমাণ করিঃ
(i) ${\left(\frac{a^q}{a^r}\right)}^p\times {\left(\frac{a^r}{a^p}\right)}^q\times {\left(\frac{a^p}{a^q}\right)}^r=1$

সমাধানঃ 

${\left(\frac{a^q}{a^r}\right)}^p\times {\left(\frac{a^r}{a^p}\right)}^q\times {\left(\frac{a^p}{a^q}\right)}^r$

=$a^{(q-r)p}\times a^{(r-p)q}\times a^{(p-q)r}$

=$a^{pq-pr}\times a^{qr-pq}\times a^{pr-qr}$

=$a^{pq-pr+qr-pq+pr-qr}$

=$a^0$

=1 [প্রমানিত]

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

সমাধানঃ 
${\left(\frac{x^m}{x^n}\right)}^{m+n}\times {\left(\frac{x^n}{x^l}\right)}^{n+l}\times {\left(\frac{x^l}{x^m}\right)}^{l+m}$
=$x^{(m-n)(m+n)}\times x^{(n-l)(n+l)}\times x^{(l-m)(l+m)}$
=$x^{m^2-n^2}\times x^{n^2-l^2}\times x^{l^2-m^2}$
=$x^{m^2-n^2+n^2-l^2+l^2-m^2}$
=$x^0$
=1 [প্রমানিত]



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

সমাধানঃ 
${\left(\frac{x^m}{x^n}\right)}^{m+n-l}\times {\left(\frac{x^n}{x^l}\right)}^{n+l-m}\times {\left(\frac{x^l}{x^m}\right)}^{l+m-n}$
=$x^{(m-n)(m+n-l)}\times x^{(n-l)(n+l-m)}\times x^{(l-m)(l+m-n)}$
=$x^{(m-n)(m+n)-(m-n)l}\times x^{(n-l)(n+l)-(n-l)m}$
        $\times x^{(l-m)(l+m)-(l-m)n}$
=$x^{m^2-n^2-ml+nl}\times x^{n^2-l^2-mn+ml}\times x^{l^2-m^2-nl+mn}$
=$x^{m^2-n^2-ml+nl+n^2-l^2-mn+ml+l^2-m^2-nl+mn}$
=$x^0$
=1 [প্রমানিত]


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

সমাধানঃ 
${\left(a^{\frac{1}{x-y}}\right)}^{\frac{1}{x-z}}\times {\left(a^{\frac{1}{y-z}}\right)}^{\frac{1}{y-x}}\times {\left(a^{\frac{1}{z-x}}\right)}^{\frac{1}{z-y}}$
=${\left(a\right)}^{\frac{1}{(x-y)(x-z)}}\times {\left(a\right)}^{\frac{1}{(y-z)(y-x)}}\times {\left(a\right)}^{\frac{1}{(z-x)(z-y)}}$
=$a^{\frac{1}{(x-y)(x-z)}+\frac{1}{(y-z)(y-x)}+\frac{1}{(z-x)(z-y)}}$
=$a^{\frac{1}{(x-y)(x-z)}-\frac{1}{(y-z)(x-y)}+\frac{1}{(x-z)(y-z)}}$
=$a^{\frac{y-z+x-z+x-y}{(x-y)(x-z)(y-z)}}$
=$a^{\frac{0}{(x-y)(x-z)(y-z)}}$
=$a^0$
=1 [প্রমানিত]
 

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

সমাধানঃ 
x+z=2y
বা, x=2y-z
বামপক্ষ

=$a^{y-z}{b}^{z-x}{c}^{x-y}$

=$a^{y-z}{b}^{z-(2y-z)}{c}^{2y-z-y}$
=$a^{y-z}{b}^{z-2y+z}{c}^{y-z}$
=$a^{y-z}{b}^{2(z-y)}{c}^{y-z}$
=$a^{y-z}(b^2{)}^{(z-y)}{c}^{y-z}$
=$a^{y-z}(ac{)}^{(z-y)}{c}^{y-z}$
=$a^{y-z}{a}^{z-y}{c}^{z-y}{c}^{y-z}$
=$a^{y-z+z-y}{c}^{z-y+y-z}$
=$a^0\times c^0$
= 1
= ডানপক্ষ 
$\therefore$ বামপক্ষ = ডানপক্ষ [প্রমানিত]


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

সমাধানঃ 
$a=x{y}^{p-1},b=x{y}^{q-1}$ এবং $c=x{y}^{r-1}$
$a^{q-r}{b}^{r-p}{c}^{p-q}$
=$(x{y}^{p-1}{)}^{(q-r)}(x{y}^{q-1})(r-p)(x{y}^{r-1}{)}^{(p-q)}$ 
=$x^{(q-r)}{y}^{(p-1)(q-r)}{x}^{(r-p)}{y}^{(q-1)(r-p)}{x}^{(p-q)}{y}^{(r-1)(p-q)}$
=$x^{q-r+r-p+p-q}{y}^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}$
=$x^0{y}^{pq-pr-q+r+qr-pq-r+p+pr-qr-p+q}$
=$1.y^0$
=1 [প্রমানিত]


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

সমাধানঃ 
ধরি, $x^{\frac{1}{a}}=y^{\frac{1}{b}}=z^{\frac{1}{c}}=k$ [যেখানে $k\ne 0,1,-1$]
$\therefore x^{\frac{1}{a}}=k$

$y^{\frac{1}{b}}=k$

$z^{\frac{1}{c}}=k$
$\therefore x=k^a,y=k^b,z=k^c$
আবার, xyz=1
বা,$k^a.k^b.k^c=1$
বা,$k^{a+b+c}=k^0$
$\therefore$ a+b+c=0


8. $a^x=b^y=c^z$ এবং abc=1 হলে, দেখাও যে, xy+yz+zx=0
সমাধানঃ 

ধরি, $a^x=b^y=c^z=k$ [যেখানে $k\ne 0,1,-1$]
$\therefore a^x=k$

$b^y=k$

$c^z=k$
$\therefore a=k^{\frac{1}{x}},b=k^{\frac{1}{y}},c=k^{\frac{1}{z}}$

আবার, abc=1
বা,$k^{\frac{1}{x}}.k^{\frac{1}{y}}.k^{\frac{1}{z}}=1$
বা,$k^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=k^0$
বা,$k^{\frac{yz+zx+xy}{xyz}}=k^0$
বা,$\frac{yz+zx+xy}{xyz}=0$
বা,yz+zx+xy=0
∴ xy+yz+zx=0 [প্রমানিত]


9. সমাধান করোঃ
(i) ${49}^x=7^3$

সমাধানঃ 
$(7^2)x=7^3$
বা,$7^{2x}=7^3$
বা,2x=3

∴ x= $\frac{3}{2}$

∴ নির্ণেয় সমাধান, $x=\frac{3}{2}$

(ii) $2^{x+2}+2^{x-1}=9$

সমাধানঃ 
$2^{x+2}+2^{x-1}=9$
বা,$2^x{.2}^2+\frac{2^x}{2}=9$
বা,$2^x(4+\frac{1}{2})=9$
বা,$2^x\left(\frac{8+1}{2}\right)=9$
বা,$2^x=9\times \frac{2}{9}$
বা,$2^x=2$

∴ x=1
∴ নির্ণেয় সমাধান, x=1


(iii) $2^{x+1}+2^{x+2}=48$

সমাধানঃ 
$2^{x+1}+2^{x+2}=48$
বা,$2^x{.2}^1+2^x{.2}^2=48$
বা,$2^x(2+4)=48$
বা,$2^x\times 6=48$
বা,$2^x=\frac{48}{6}$
বা,$2^x=8$
বা,$2^x=2^3$
বা,x=3
∴ নির্ণেয় সমাধান, x=3




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

সমাধানঃ 
$2^{4x}{.4}^{3x-1}=\frac{4^{2x}}{2^{3x}}$
বা,$2^{4x}{.4}^{3x-1}.\frac{2^{3x}}{4^{2x}}=1$
বা,$2^{4x+3x}{.4}^{3x-1-2x}=1$
বা,$2^{7x}.(2^2{)}^{x-1}=1$
বা,$2^{7x}{.2}^{2x-2}=1$
বা,$2^{7x+2x-2}=2^0$
বা,$2^{9x-2}=2^0$
বা,9x-2=0
বা,x=$\frac{2}{9}$
∴ নির্ণেয় সমাধান, x=$\frac{2}{9}$


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

সমাধানঃ 
$9\times {81}^x={27}^{2-x}$
বা,$3^2\times {\left(3^4\right)}^x={\left(3^3\right)}^{2-x}$
বা,$3^2\times 3^{4x}=3^{6-3x}$
বা,$\frac{3^{2+4x}}{3^{6-3x}}=1$
বা,$3^{(2+4x)-(6-3x)}=1$
বা,$3^{2+4x-6+3x}=3^0$
বা,$3^{7x-4}=3^0$
বা,7x-4=0
বা,7x=4
বা,x=$\frac{4}{7}$

∴ নির্ণেয় সমাধান, x=$\frac{4}{7}$

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

সমাধানঃ 
$2^{5x+4}+2^9=2^{10}$
বা,$2^{5x}{.2}^4+2^5{.2}^4=210$
বা,$2^4(25x+25)=210$
বা,$(25x+25)=\frac{2^{10}}{2^4}$
বা,$2^{5x}=2^6-2^5$
বা,$2^{5x}=64-32$
বা,$2^{5x}=32$
বা,$2^{5x}=2^5$
বা,5x=5
বা,x=1
∴ নির্ণেয় সমাধান, x=1


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

সমাধানঃ 
$6^{2x+4}=3^{3x}{.2}^{x+8}$
বা,$(2.3{)}^{2x+4}=3^{3x}{.2}^{x+8}$
বা,$2^{2x+4}{.3}^{2x+4}=3^{3x}{.2}^{x+8}$
বা,$\frac{2^{2x+4}{.3}^{2x+4}}{3^{3x}{.2}^{x+8}}=1$
বা,$\frac{2^{2x+4-x-8}}{3^{3x-2x-4}}=1$
বা,$\frac{2^{x-4}}{3^{x-4}}=1$
বা,${\left(\frac{2}{3}\right)}^{x-4}={\left(\frac{2}{3}\right)}^0$
বা,x-4=0
বা, x=4
∴ নির্ণেয় সমাধান, x=4


(viii) $2^{(x^2-4)}=3^{(x^2-4)}$

সমাধানঃ 
$2^{(x^2-4)}=3^{(x^2-4)}$
বা,$\frac{2^{(x^2-4)}}{3^{(x^2-4)}}=1$
বা,${\left(\frac{2}{3}\right)}^{(x^2-4)}={\left(\frac{2}{3}\right)}^0$
বা, $x^2-4=0$
বা,$x^2=4$
∴ x=±2
∴ নির্ণেয় সমাধান, x=±2


10. বহু বিকল্পীয় প্রশ্ন(M.C.Q): 
(i) $(0.243{)}^{0.2}\times (10{)}^{0.6}$ এর মান
a. 0.3 b. 3     c. 0.9     d. 9
সমাধানঃ 

$(0.243{)}^{0.2}\times (10{)}^{0.6}$
=${\left(\frac{243}{1000}\right)}^{\frac{1}{5}}\times {(10)}^{\frac{3}{5}}$
=${\left(\frac{3^5}{{10}^3}\right)}^{\frac{1}{5}}\times {(10)}^{\frac{3}{5}}$
=$\frac{3}{{10}^{\frac{3}{5}}}\times {10}^{\frac{3}{5}}$
= 3
উত্তরঃ b. 3


(ii)$2^{\frac{1}{2}}\times 2^{-\frac{1}{2}}\times {(16)}^{\frac{1}{2}}$ এর মান
a. 1     b. 2 c. 4     d. $frac12$

সমাধানঃ 
$2^{\frac{1}{2}}\times 2^{-\frac{1}{2}}\times {\left(16\right)}^{\frac{1}{2}}$
=$2^{\frac{1}{2}-\frac{1}{2}}\times {(2^4)}^{\frac{1}{2}}$
= $2^0\times 2^{4\times \frac{1}{2}}$
= $1\times 2^2$
= 4
উত্তরঃ c. 4

 
(iii) $4^x=8^3$ হলে, x এর মান
a. $\frac{3}{2}$    b. $\frac{9}{2}$     c. 3     d. 9

সমাধানঃ 
$4^x=8^3$
বা,$(2^2{)}^x=(2^3{)}^3$
বা,$2^{2x}=2^9$
বা,2x=9
বা,x=$\frac{9}{2}$
উত্তরঃ b. $\frac{9}{2}$


(iv) ${20}^{-x}=\frac{1}{7}$ হলে, $(20{)}^{2x}$ এর মান
a.$\frac{1}{49}$ b. 7         c. 49     d. 1

সমাধানঃ 
${20}^{-x}=\frac{1}{7}$
বা,$\frac{1}{{20}^x}=\frac{1}{7}$
বা${20}^x=7$
বা,$({20}^x{)}^2=7^2$
বা,${20}^{2x}=49$

উত্তরঃ c. 49
 
(v)$4\times 5^x=500$ হলে, $x^x$ এর মান
a. 8     b. 1 c. 64 d. 27

সমাধানঃ 
$4\times 5^x=500$
বা,$5^x=125$
বা, $5^x=5^3$
বা,x=3
∴$x^x=3^3=27$

উত্তরঃ d. 27

11. সংক্ষিপ্ত প্রশ্নঃ
(i) $(27{)}^x=(81{)}^y$ হলে x:y কত হয় লিখি।
সমাধানঃ 
$(27{)}^x=(81{)}^y$
বা,$(3^3{)}^x=(3^4{)}^y$
বা,$3^{3x}=3^{4y}$
বা,3x=4y
বা,$\frac{x}{y}=\frac{4}{3}$
∴ x:y=4:3


(ii) $(5^5+0.01{)}^2-(5^5-0.01{)}^2=5^x$ হলে, x এর মান কত হবে হিসাব করে লিখি। 

সমাধানঃ 
$(5^5+0.01{)}^2-(5^5-0.01{)}^2=5^x$
বা,$4\times 5^5\times 0.01=5^x$
বা,$4\times \frac{1}{100}=\frac{5^x}{5^5}$
বা,$\frac{1}{25}=5^{x-5}$
বা,$5^{x-5}=\frac{1}{5^2}$
বা, $5^{x-5}=5^{-2}$
বা, x-5=-2
বা, x=-2+5
∴ x=3


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

সমাধানঃ 
$3\times {27}^x=9^{x+4}$
বা,$3\times (3^3{)}^x=(32)x+4$
বা,$3\times 3^{3x}=3^{2x+8}$
বা,$3^{1+3x}=3^{2x+8}$
বা,1+3x=2x+8
বা,3x-2x=8-1
∴ x=7


(iv) $\sqrt[3]{{\left(\frac{1}{64}\right)}^{\frac{1}{2}}}$ হলে, x এর মান হিসাব করে লিখি। 

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


(v) $3^{3^3}$এবং $(3^3{)}^3$ এর মধ্যে কোনটি বৃহত্তর যুক্তিসহ লিখি।

সমাধানঃ 
 $3^{3^3}=3^{27}$
এবং $(3^3{)}^3=3^9$
যেহেতু 27> 9
∴$3^{3^3}>(3^3{)}^3$
∴ বৃহত্তর সংখ্যাটি হল $3^{3^3}$