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SSC বোর্ড পরীক্ষার প্রশ্নের সমাধান

অনুশীলনী-৫.৩   

১. {3^{x + 2}} = 81

সমাধান:  {3^{x + 2}} = 81

\Rightarrow {3^{x + 2}} = {3^4}

\Rightarrow x + 2 = 4 \because \left[ {{a^m} = {a^n}} \right]  হলে, \left[ {m = n} \right]

\Rightarrow x = 4 - 2

\therefore x = 2

\therefore নির্ণেয় সমাধান: x = 2

২. {5^{3x - 7}} = {3^{3x - 7}}

সমাধান:  {5^{3x - 7}} = {3^{3x - 7}}

\Rightarrow {{{5^{3x}}} \over {{5^7}}} = {{{3^{3x}}} \over {{3^7}}}  \because \left[ {{a^{m - n}} = {{{a^m}} \over {{a^n}}}} \right]

\Rightarrow {{{5^{3x}}} \over {{3^{3x}}}} = {{{5^7}} \over {{3^7}}}

\Rightarrow {\left( {{5 \over 3}} \right)^{3x}} = {\left( {{5 \over 3}} \right)^7}  \because \left[ {{{{a^n}} \over {{b^n}}} = {{\left( {{a \over b}} \right)}^n}} \right]

\Rightarrow 3x = 7

\therefore x = {7 \over 3}

\therefore নির্ণেয় সমাধান:  x = {7 \over 3}

৩.  {2^{x - 4}} = 4{a^{x - 6}},\left( {a > 0,a \ne 2} \right)

সমাধান:  {2^{x - 4}} = 4{a^{x - 6}}

\Rightarrow {{{2^{x - 4}}} \over 4} = {a^{x - 6}}

\Rightarrow {{{2^{x - 4}}} \over {{2^2}}} = {a^{x - 6}}

\Rightarrow {2^{x - 4 - 2}} = {a^{x - 6}}   \because \left[ {{{{a^m}} \over {{a^n}}} = {a^{m - n}}} \right]

\Rightarrow {2^{x - 6}} = {a^{x - 6}}

\Rightarrow {{{2^{x - 6}}} \over {{a^{x - 6}}}} = 1

\Rightarrow {\left( {{2 \over a}} \right)^{x - 6}} = 1   \because \left[ {{{{a^m}} \over {{b^m}}} = {{\left( {{a \over b}} \right)}^m}} \right]

\Rightarrow {\left( {{2 \over a}} \right)^{x - 6}} = {\left( {{2 \over a}} \right)^0}  \because \left[ {{p^0} = 1} \right]

\Rightarrow x - 6 = 0

\therefore x = 6

\therefore নির্ণেয় সমাধান: x = 6

৪. {\left( {\sqrt 3 } \right)^{x + 5}} = {\left( {\root 3 \of 3 } \right)^{2x + 5}}

সমাধান:  {\left( {\sqrt 3 } \right)^{x + 5}} = {\left( {\root 3 \of 3 } \right)^{2x + 5}}

\Rightarrow {\left( {{3^{{1 \over 2}}}} \right)^{x + 5}} = {\left( {{3^{{1 \over 3}}}} \right)^{2x + 5}}

\Rightarrow {3^{{{x + 5} \over 2}}} = {3^{{{2x + 5} \over 3}}}  \because \left[ {{{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]

\Rightarrow {{x + 5} \over 2} = {{2x + 5} \over 3}

\Rightarrow 2\left( {2x + 5} \right) = 3\left( {x + 5} \right)

\Rightarrow 4x + 10 = 3x + 15

\Rightarrow 4x - 3x = 15 - 10

\therefore x = 5

\therefore নির্ণেয় সমাধান: x = 5

৫. {\left( {\root 5 \of 4 } \right)^{4x + 7}} = {\left( {\root {11} \of {64} } \right)^{2x + 7}}

সমাধান:  {\left( {\root 5 \of 4 } \right)^{4x + 7}} = {\left( {\root {11} \of {64} } \right)^{2x + 7}}

\Rightarrow {\left( {\root 5 \of 4 } \right)^{4x + 7}} = {\left( {\root {11} \of {{4^3}} } \right)^{2x + 7}}

\Rightarrow {\left( {{4^{{1 \over 5}}}} \right)^{4x + 7}} = {\left( {{4^{{3 \over {11}}}}} \right)^{2x + 7}}    \because \left[ {\root n \of {{a^m}}  = {a^{{m \over n}}}} \right]

\Rightarrow {4^{{1 \over 5}}}^{\left( {4x + 7} \right)} = {4^{{3 \over {11}}\left( {2x + 7} \right)}}

\because \left[ {{{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]

\Rightarrow {1 \over 5}\left( {4x + 7} \right) = {3 \over {11}}\left( {2x + 7} \right)

\Rightarrow 11\left( {4x + 7} \right) = 5 \times 3\left( {2x + 7} \right)

\Rightarrow 44x + 77 = 30x + 105

\Rightarrow 44x - 30x = 105 - 77

\Rightarrow 14x = 28

\Rightarrow x = {{28} \over {14}}

\therefore x = 2

\therefore নির্ণেয় সমাধান: x = 2

৬. {{{3^{3x - 4}}.{a^{2x - 5}}} \over {{3^{x + 1}}}} = {a^{2x - 5}}  \left( {a > 0} \right)

সমাধান: {{{3^{3x - 4}}.{a^{2x - 5}}} \over {{3^{x + 1}}}} = {a^{2x - 5}}

\Rightarrow {3^{3x - 4 - x - 1}} = {{{a^{2x - 5}}} \over {{a^{2x - 5}}}}

\Rightarrow {3^{2x - 5}} = 1

\Rightarrow {3^{2x - 5}} = {3^0}  \because \left[ {{a^0} = 1} \right]

\Rightarrow 2x - 5 = 0

\Rightarrow 2x = 5

\therefore x = {5 \over 2}

\therefore নির্ণেয় সমাধান: x = {5 \over 2}

৭.   {{{5^{2x}}.{b^{x - 3}}} \over {{5^{x + 3}}}} = {a^{x - 3}} \left( {a,b > 0,5b \ne a} \right)

সমাধান:  {{{5^{2x}}.{b^{x - 3}}} \over {{5^{x + 3}}}} = {a^{x - 3}}

\Rightarrow {{{5^{2x}}} \over {{5^{x + 3}}}} = {{{a^{x - 3}}} \over {{b^{x - 3}}}}

\Rightarrow {5^{2x - x - 3}} = {\left( {{a \over b}} \right)^{x - 3}}

\Rightarrow {5^{x - 3}} = {\left( {{a \over b}} \right)^{x - 3}}

\Rightarrow {\left( {{5 \over {{a \over b}}}} \right)^{x - 3}} = {\left( {{5 \over {{a \over b}}}} \right)^0}

\Rightarrow x - 3 = 0

\therefore x = 3

\therefore নির্ণেয় সমাধান: x = 3

৮. {4^{x + 2}} = {2^{2x + 1}} + 14

সমাধান: {4^{x + 2}} = {2^{2x + 1}} + 14

\Rightarrow {4^2} \times {4^x} = {2^{2x}} \times 2 + 14

\Rightarrow 16 \times {4^x} = {\left( {{2^2}} \right)^x} \times 2 + 14

\Rightarrow 16 \times {4^x} = {4^x} \times 2 + 14

\Rightarrow 16 \times {4^x} - {4^x} \times 2 = 14

\Rightarrow {4^x}\left( {16 - 2} \right) = 14

\Rightarrow {4^x} \times 14 = 14

\Rightarrow {4^x} = {{14} \over {14}}

\Rightarrow {4^x} = 1

\Rightarrow {4^x} = {4^0}    \because \left[ {{a^0} = 1} \right]

\Rightarrow x = 0

\therefore নির্ণেয় সমাধান: x = 0

৯. {5^x} + {5^{2 - x}} = 26

সমাধান:  {5^x} + {5^{2 - x}} = 26

\Rightarrow {5^x} + {{{5^2}} \over {{5^x}}} \because \left[ {{a^{m - n}} = {{{a^m}} \over {{a^n}}}} \right]

\Rightarrow {5^x} \times {5^x} + {5^2} = {26.5^x}

\Rightarrow {5^x} \times {5^x} + {5^2} = {26.5^x}

\Rightarrow {\left( {{5^x}} \right)^2} - 26 \times {5^x} + 25 = 0

\Rightarrow {a^2} - 26a + 25 = 0  \left[ {{5^x} = a} \right] ধরে

\Rightarrow {a^2} - 25a - a + 25 = 0

\Rightarrow a\left( {a - 25} \right) - 1\left( {a - 25} \right) = 0

\Rightarrow \left( {a - 25} \right)\left( {a - 1} \right) = 0

হয়, a - 25 = 0

\therefore a = 25

a =25 হলে,

{5^x} = 25

\Rightarrow {5^x} = {5^2}

\therefore x = 2

অথবা, a - 1 = 0

\therefore a = 1

a - 1 হলে,

{5^x} = 1

\Rightarrow {5^x} = {5^0}

\therefore x = 0

\therefore নির্ণেয় সমাধান: x = 0,2

১০.  3({9^x} - {4.3^{x - 1}}) + 1 = 0

সমাধান: 3({9^x} - {4.3^{x - 1}}) + 1 = 0

বা, {3.9^x} - {4.3.3^{x - 1}} + 1 = 0

বা, 3.{({3^2})^x} - {4.3^{x - 1 + 1}} + 1 = 0   \left[ {\because {a^m}.{a^n} = {a^{m + n}}} \right]

বা, 3.{({3^x})^2} - {4.3^x} + 1 = 0    \left[ {{{\left( {{a^m}} \right)}^n} = {a^{mn}} = {a^{nm}} = {{\left( {{a^n}} \right)}^m}} \right]

বা, 3{a^2} - 4a + 1 = 0    \left[ {{3^x} = a} \right] ধরে

বা, 3{a^2} - 3a - a + 1 = 0

বা, 3a(a - 1) - 1(a - 1) = 0

বা, (3a - 1)(a - 1) = 0

হয়, 3a - 1 = 0                              অথবা, a - 1 = 0

\therefore a = \frac{1}{3}          \therefore a = 1

a = \frac{1}{3} হলে,                 a = 1 হলে ,

{3^x} = \frac{1}{3}                বা, {3^x} = 1

বা, 3x = {3^{ - 1}}                   বা, {3^x} = {3^0}

\therefore x =  - 1                    \therefore x = 0

\therefore নির্ণেয় সমাধান: x = 0, - 1

১১. {4^{1 + x}} + {4^{1 - x}} = 10

সমাধান: {4^{1 + x}} + {4^{1 - x}} = 10

{4.4^x} + \frac{4}{{{4^x}}} = 10   \left[ {\because {a^{m + n}} = {a^m}{a^n};{a^{m - n}} = \frac{{{a^m}}}{{{a^n}}}} \right]

\Rightarrow {4.4^x}{.4^x} + 4 = {10.4^x}

\Rightarrow 4{\left( {{4^x}} \right)^2} - {10.4^x} + 4 = 0

\Rightarrow 4{a^2} - 10a + 4 = 0   \left[ {{4^x} = a} \right] ধরে

\Rightarrow 4{a^2} - 8a - 2a + 4 = 0

\Rightarrow 4a\left( {a - 2} \right) - 2\left( {a - 2} \right) = 0

\Rightarrow \left( {a - 2} \right)\left( {4a - 2} \right) = 0

হয়, a - 2 = 0

\therefore a = 2

অথবা, 4a - 2 = 0

\Rightarrow 4a = 2

\Rightarrow a = \frac{2}{4}

\therefore a = \frac{1}{2}

এখন, a = 2 হলে,  {4^x} = 2

\Rightarrow {\left( 4 \right)^x} = {4^{\frac{1}{2}}}  \left[ {\because \sqrt 4  = {4^{\frac{1}{2}}} = 2} \right]

\therefore x = \frac{1}{2}

আবার, a = \frac{1}{2} হলে, {4^x} = \frac{1}{2}

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

\therefore x =  - \frac{1}{2}

\therefore নির্ণেয় সমাধান: x = \frac{1}{2}, - \frac{1}{2}

১২. {22^x} - {3.2^{x + 2}} =  - 32

সমাধান: {22^x} - {3.2^{x + 2}} =  - 32

\Rightarrow {\left( {{2^x}} \right)^2} - {3.2^2}{.2^x} =  - 32  \left[ {\because {a^{mn}} = {{\left( {{a^m}} \right)}^n};{a^{m + n}} = {a^m}.{a^n}} \right]

\Rightarrow {\left( {{2^x}} \right)^2} - {12.2^x} + 32 = 0

\Rightarrow {a^2} - 12a + 32 = 0   \left[ {{2^x} = a} \right] ধরে

\Rightarrow {a^2} - 8a - 4a + 32 = 0

\Rightarrow a\left( {a - 8} \right) - 4\left( {a - 8} \right) = 0

\Rightarrow \left( {a - 4} \right)\left( {a - 8} \right) = 0

হয়, a - 4 = 0

\therefore a = 4

অথবা, a - 8 = 0

\therefore a = 8

এখন, a = 4 হলে,

{2^x} = 4

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

\therefore x = 2

আবার, a = 8

{2^x} = 8

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

\therefore x = 3

\therefore নির্ণেয় সমাধান: x = 2,3

১. {\left( {\sqrt 5 } \right)^{x + 3}} = 125 এর সমাধান কত? [ঢা. বো. ১৭]

(a) – 3               (b) \sqrt 5             (c) 3                   (d) 5

উত্তর:  (c) 3

২. যদি \root 3 \of {{x^2}}  = {({x^a}\sqrt {{x^a}} )^b} হয়, তাহলে ab এর মান কত? [ঢা. বো. ১৭]

(a) 1               (b) \frac{2}{3}             (c) \frac{4}{9}         (d) \frac{2}{9}

উত্তর:  (c) \frac{4}{9}

৩. যদি {(16)^{\frac{1}{p}}} = {(64)^{\frac{1}{q}}} হয়, তবে \frac{p}{q} এর মান কত হবে? [দি.বো. ১৭]

(a) \frac{1}{3}                 (b) \frac{2}{3}

(c) \frac{3}{2}                  (d) \frac{8}{3}

উত্তর: (b) \frac{2}{3}

৪. {3^{y + 8}} = {9^{2y + 1}} হলে, y এর মান কত? [কু. বো. ১৭]

(a) 6               (b) \frac{{10}}{3}             (c) \frac{{7}}{3}                  (d) 2

উত্তর: (d) 2

৫. {9^{2x}} = {3^{x + 1}} হলে, x= কত? [চ. বো. ১৭]

(a) - \frac{1}{3}        (b) 0        (c) \frac{1}{3}            (d) \frac{1}{5}

উত্তর: (c) \frac{1}{3}  

৬. {3.27^x} = {9^{x + 4}} এর সঠিক সমাধান কোনটি? [য. বো. ১৭]

(a) 6               (b) 7              (c) 8                   (d) 9

উত্তর: (b) 7

৭. {3^{mx - 1}} = 3{a^{mx - 2}};[a > 0,a \ne 3,m \ne 0] হলে, x এর মান কত? [ঢা. বো. ১৬]

(a) \frac{m}{2}         (b) \frac{2}{m}           (c) 2m              (d) {2^m}

উত্তর: (b) \frac{2}{m}  

৮. {3^{ax - 1}} = 3{b^{ax - 2}} এর সমাধান কোনটি? [চ. বো. ১৬]

(a) \frac{a}{2}                 (b) \frac{{ - 2}}{a}

(c) \frac{1}{a}                  (d) \frac{2}{a}

উত্তর: (d) \frac{2}{a}

৯. {2^{x + 7}} = {4^{x + 2}} সমীকরণটির সমাধান নিচের কোনটি? [রা. বো. ১৬; সি. বো. ১৫]

(a) 5               (b) 4              (c) 3                   (d) 2

উত্তর: (c) 3 

১০. {a^x} = {a^y} হলে, x = y হবে কোন শর্তে? [সি. বো. ১৫]

(a) a > 0                             (b) a < 0,a \ne 1

(c) a < 0                             (d) a > 0,a \ne 1

উত্তর: (d) a > 0,a \ne 1

১১. যদি {4^x} = 16 হয়, তবে x =  কত? [চ. বো. ১৫]

(a) 2               (b) 4              (c) 8                   (d) 16

উত্তর: (a) 2

১২. {3^{3x}} = \frac{1}{3} হলে, x– এর মান কত? [কু. বো. ১৫]

(a) – 3               (b) \frac{{ - 1}}{3}             (c) \frac{1}{3}                  (d) 3

উত্তর: (b) \frac{{ - 1}}{3}

১৩. {9^{2x}} = {3^{5x - 2}} সমীকরণটির সমাধান কোনটি? [য. বো. ১৫]

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

উত্তর: (d) 2

১৪. {5^{3x - 7}} = {3^{3x - 7}} সমীকরণে x এর মান কত? [চ. বো. ১৫]

(a) \frac{1}{3}                 (b) \frac{2}{3}

(c) \frac{5}{3}                  (d) \frac{7}{3}

উত্তর: (d) \frac{7}{3}

১৫. {9^{x + 5}} = {81^{x + 1}} হলে, x এর মান কত? [ন.প্র.চ.বো.]

(a) – 6               (b) – 3              (c) 3                   (d) 6

উত্তর: (c) 3