SSC general math exponent (সূচক) chapter 4.1 solution
প্রশ্ন ১: \frac{3^3.3^5}{3^6}
সমাধান: \frac{3^3.3^5}{3^6}
= \frac{3^{3+5}}{3^6} [⸪ a^m × a^n = a^{m + n}]
= 3^{8-6} [⸪ \frac{a^m} {a^n} = a^{m - n}]
= 32
= 9
প্রশ্ন ২: \frac{5^3.8}{2^4.125}
সমাধান: \frac{5^3.8}{2^4.125}
= \frac{5^3. 2^3}{2^4.5^3}
= 5^{3 - 3} × 2^{3 - 4} [⸪ \frac{a^m} {a^n} = a^{m - n}]
= 5^0 × 2^{-1}
= 1 × \frac{1}{2}
= \frac{1}{2}
প্রশ্ন ৩: \frac{7^3 × 3^{-3}}{3 × 3^{-4}}
সমাধান: \frac{7^3 × 3^{-3}}{3 × 3^{-4}}
= \frac{7^{3 - 3}}{3^{1 - 4}} [⸪ a^m × a^{-n} = a^{m - n}]
= \frac{1}{3^{-3}} [⸪ \frac{a^m} {a^n} = a^{m - n}]
= 33
= 27
প্রশ্ন ৪: \frac{\sqrt[3]{7^2}.\sqrt[3]7}{\sqrt7}\\
সমাধান: \frac{\sqrt[3]{7^2}.\sqrt[3]7}{\sqrt7}\\
=\frac{\left(7^2\right)^{\displaystyle\frac13}. \left(7\right)^{\displaystyle\frac17}}{7^{\displaystyle\frac12}}\\
=\frac{7^{\displaystyle\frac23}.7^{\displaystyle\frac13}}{7^{\displaystyle\frac12}}\\
=\frac{7^{{\displaystyle\frac23}+{\displaystyle\frac13}}}{7^{\displaystyle\frac12}}\\
=\;\frac{7^{\displaystyle\frac{2+3}3}}{7^{\displaystyle\frac12}}\\
=\frac{7^{\displaystyle\frac33}}{7^{\displaystyle\frac12}}\\
=\frac{7^1}{7^{\displaystyle\frac12}}\\
=7^{1-\frac12}\\
=7^\frac12\\
=\sqrt7 
প্রশ্ন:৫।। \left(2a^{-1}+3b^{-1}\right)^{-1}
সমাধান:\left(2a^{-1}+3b^{-1}\right)^{-1}\\=\left(2\times\frac1a+3\times\frac1b\right)^{-1}\\=\left(\frac2a+\frac3b\right)^{-1}\\=\left(\frac{2b+3a}{ab}\right)^{-1}\\=\frac1{\displaystyle\frac{2b+3a}{ab}}\\=1\times\frac{ab}{3a+2b}\\=\frac{ab}{3a+2b}
প্রশ্ন: ৬।। \left(\frac{a^2b^{-1}}{a^{-2}b}\right)^2
সমাধান: \left(\frac{a^2b^{-1}}{a^{-2}b}\right)^2\\\\=\left(\frac{a^2}{a^{-2}}\times\frac{b^{-1}}b\right)^2\\\\=\left(a^{2-(-2)}\times b^{-1-1}\right)\\\\=\left(a^{2+2}\times b^{-2}\right)^2\\\\=\left(\frac{a^4}{b^2}\right)^2\\\\=\frac{\left(a^4\right)^2}{\left(b^2\right)^2}\\\\=\frac{a^{4\times2}}{b^{2\times2}}\\\\=\frac{a^8}{b^4}
প্রশ্ন: ৭।। \sqrt{x^{-1}y}.\sqrt{y^{-1}z}.\sqrt{z^{-1}x}</span> <span style="font-size: 18pt;">
\sqrt{x^{-1}y}.\sqrt{y^{-1}z}.\sqrt{z^{-1}x}\\\\=\sqrt{x^{-1}y.y^{-1}z.z^{-1}x}\\\\=\sqrt{x^{-1+1}.y^{1-1}.z^{1-1}}\\\\=\sqrt{x^0.y^0.z^0}\\\\=\sqrt{1.1.1}\\\\=\sqrt1\\\\=1\\
প্রশ্ন: ৮।। \frac{2^{n+4}-4.2^{n+1}}{2^{n+2}\div2}
\frac{2^{n+4}-4.2^{n+1}}{2^{n+2}\div2}\\\\=\frac{2^n.2^4-2^2.2^n.2^1}{2^n.2^2\div2}\\\\=\frac{2^n.2^4-2^{2+1}.2^n}{2^n.2^{2-1}}\\\\=\frac{2^n.2^4-2^3.2^n}{2^n.2^1}\\\\=\frac{2^n.2\left(2^3-2^2\right)}{2^n.2}\\\\=2^3-2^2\\\\=8-4\\\\=4
প্রশ্ন ৯।
\frac{3^{m+1}}{(3^m)^{m-1}} \div \frac{9^{m+1}}{(3^{m-1})^{m+1}}
সমাধান:
\frac{3^{m+1}}{(3^m)^{m-1}} \div \frac{9^{m+1}}{(3^{m-1})^{m+1}}
= \frac{3^{m+1}}{(3^m)^{m-1}} \div \frac{(3^2)^{m+1}}{(3^{m-1})^{m+1}}
= \frac{3^{m+1}}{3^{m^2-m}} \div \frac{3^{2m+2}}{3^{m^2-1}}
= \frac{3^{m+1}}{3^{m^2-m}} \cdot \frac{3^{m^2-1}}{3^{2m+2}}
= 3^{m+1-(m^2-m)} \cdot 3^{m^2-1-(2m+2)}
= 3^{m+1-m^2+m} \cdot 3^{m^2-2m-3}
= 3^{(m+1-m^2+m) + (m^2-2m-3)}
= 3^{m+1-m^2+m + m^2-2m-3}
= 3^{-2}
= \frac{1}{3^2}
= \frac{1}{9}; (\text{Ans.}).
প্রশ্ন ১০। \quad \frac{4^n - 1}{2^n - 1} = 2^n + 1
সমাধান: \quad \text{বামপক্ষ } = \frac{4^n - 1}{2^n - 1}
= \frac{(2^2)^n - 1}{2^n - 1}
= \frac{(2^n)^2 - (1)^2}{2^n - 1}
= \frac{(2^n + 1)(2^n - 1)}{2^n - 1}
= 2^n + 1 \quad = \quad \text{ডানপক্ষ}
\therefore \frac{4^n - 1}{2^n - 1} = 2^n + 1 \quad ({প্রমাণিত})
প্রশ্ন:১১।। \frac{2^{p+1} \cdot 3^{2p-q} \cdot 5^{p+q} \cdot 6^{q}}{6^{p} \cdot 10^{q+2} \cdot 15^{p}}
সমাধান: \frac{2^{p+1} \cdot 3^{2p-q} \cdot 5^{p+q} \cdot 6^{q}}{6^{p} \cdot 10^{q+2} \cdot 15^{p}}
= \frac{2^{p+1} \cdot 3^{2p-q} \cdot 5^{p+q} \cdot (2 \cdot 3)^{q}}{(2 \cdot 3)^{p} \cdot (2 \cdot 5)^{q+2} \cdot (3 \cdot 5)^{p}}
= \frac{2^{p+1} \cdot 3^{2p-q} \cdot 5^{p+q} \cdot 2^{q} \cdot 3^{q}}{2^{p} \cdot 3^{p} \cdot 5^{q+2} \cdot 3^{p} \cdot 5^{p}}
= \frac{2^{p+1} \cdot 3^{2p-q} \cdot 5^{p+q} \cdot 6^{q}}{6^{p} \cdot 10^{q+2} \cdot 15^{p}}
= \frac{2^{p+1} \cdot 3^{2p-q} \cdot 5^{p+q} \cdot 3^{q} \cdot 2^{q}}{3^{q} \cdot 2^{p} \cdot 5^{q+2} \cdot 2^{q+2} \cdot 3^{p} \cdot 5^{p}}
= \frac{2^{p+1+q} \cdot 3^{2p-q+q} \cdot 5^{p+q}}{2^{p+q+2} \cdot 3^{p+q} \cdot 5^{p+q+2}}
= \frac{2^{p+q+1-p-q-2} \cdot 3^{2p-2p} \cdot 5^{p+q-p-q-2}}{1}
= 2^{-1} \cdot 3^{0} \cdot 5^{-2}
= \frac{1}{2} \cdot 1 \cdot \frac{1}{25}
= \frac{1}{50}
অতএব:
\frac{2^{p+1} \cdot 3^{2p-q} \cdot 5^{p+q} \cdot 6^{q}}{6^{p} \cdot 10^{q+2} \cdot 15^{p}} = [latex]\frac{1}{50} \quad \text{(প্রমাণিত)}
প্রশ্ন:১২
\left(\frac{a^l}{a^m}\right)^n \cdot \left(\frac{a^m}{a^n}\right)^l \cdot \left(\frac{a^n}{a^l}\right)^m = 1
সমাধান:
বামপক্ষ:
\left(\frac{a^l}{a^m}\right)^n \cdot \left(\frac{a^m}{a^n}\right)^l \cdot \left(\frac{a^n}{a^l}\right)^m
= \left(a^{l-m}\right)^n \cdot \left(a^{m-n}\right)^l \cdot \left(a^{n-l}\right)^m
= a^{n(l-m)} \cdot a^{l(m-n)} \cdot a^{m(n-l)}
= a^{ln-mn} \cdot a^{ml-nl} \cdot a^{mn-lm}
= a^{ln-mn+ml-nl+mn-lm}
= a^0 = 1
অতএব:
\left(\frac{a^l}{a^m}\right)^n \cdot \left(\frac{a^m}{a^n}\right)^l \cdot \left(\frac{a^n}{a^l}\right)^m = 1 \quad \text{(প্রমাণিত)}
প্রশ্ন: ১৩
\frac{a^{p+q}}{a^{2r}}\times\frac{a^{q+r}}{a^{2p}}\times\frac{a^{r+p}}{a^{2q}}=1
সমাধান:
বামপক্ষ: \frac{a^{p+q}}{a^{2r}}\times\frac{a^{q+r}}{a^{2p}}\times\frac{a^{r+p}}{a^{2q}}\\
=\left(a^{p+q-2r}\right)\times\left(a^{q+r-2p}\right)\times\left(a^{r+p-2q}\right)\\
=a^{{}^{p+q-2r+q+r-2p+r+p-2q}}\\
=a^{{}^{2p-2p+2q-2q+2r-2r}}\\
=a^0\\
=1
So, \frac{a^{p+q}}{a^{2r}}\times\frac{a^{q+r}}{a^{2p}}\times\frac{a^{r+p}}{a^{2q}}=1 (Proved)
প্রশ্ন: ১৪
\left(\frac{x^a}{x^b}\right)^\frac1{ab}\times\left(\frac{x^b}{x^c}\right)^\frac1{bc}\times\left(\frac{x^c}{x^a}\right)^\frac1{ca}=1
সমাধান:
বামপক্ষ: \left(\frac{x^a}{x^b}\right)^\frac1{ab}\times\left(\frac{x^b}{x^c}\right)^\frac1{bc}\times\left(\frac{x^c}{x^a}\right)^\frac1{ca}
= \left(\frac{x^a}{x^b}\right)^\frac1{ab}\times\left(\frac{x^b}{x^c}\right)^\frac1{bc}\times\left(\frac{x^c}{x^a}\right)^\frac1{ca}\\
= \left(\frac{x^a}{x^b}\right)^\frac1{ab}\times\left(\frac{x^b}{x^c}\right)^\frac1{bc}\times\left(\frac{x^c}{x^a}\right)^\frac1{ca}\\
= \left(x^{a-b}\right)^\frac1{ab}\times\left(x^{b-c}\right)^\frac1{bc}\times\left(x^{c-a}\right)^\frac1{ca}\\
= x^\frac{a-b}{ab}\times x^\frac{b-c}{bc}\times x^\frac{c-a}{ca}\\=x^\frac{ac-bc+ab-ac+bc-ab}{abc}\\
= x^\frac0{abc}\\
= x^0\\
= 1
প্রশ্ন: ১৫
\left(\frac{x^p}{x^q}\right)^{p+q-r}\times\left(\frac{x^q}{x^r}\right)^{q+r-p}\times\left(\frac{x^r}{x^p}\right)^{r+p-q}=1
সমাধান:
বামপক্ষ: \left(\frac{x^p}{x^q}\right)^{p+q-r}\times\left(\frac{x^q}{x^r}\right)^{q+r-p}\times\left(\frac{x^r}{x^p}\right)^{r+p-q}\\
=\left(x^{p-q}\right)^{p+q-r}\times\left(x^{q-r}\right)^{q+r-p}\times\left(x^{r-p}\right)^{r+p-q}\\
=x^{\left(p-q\right)\left(p+q-r\right)}\times x^{\left(q-r\right)\left(q+r-p\right)}\times x^{\left(r-p\right)\left(r+p-q\right)}\\
=x^{p^2+pq-pr-pq-q^2+qr}\times x^{q^2+qr-pq-qr-r^2+pr}\times x^{r^2+pr-qr-pr-p^2+pq}\\
=x^{{}^{p^2+pq-pr-pq-q^2+qr+q^2+qr-pq-qr-r^2+pr+r^2+pr-qr-pr-p^2+pq}}\\
=x^0\\
=1
প্রশ্ন: ১৮।। যদি a^x=b,b^y=c,c^z=a হয়, তবে দেখাও যে, xyz = 1.
সমাধান:
দেওয়া আছে,
\left(c^z\right)^x=b\\
or,c^{zx}=b\\
or,\left(b^y\right)^{zx}=b\\
or,b^{xyz}=b^1\\
or,xyz=1\\ [⸪a^x = a^y হলে x = y]
সমাধান কর: (১৯ - ২২)
প্রশ্ন:১৯।। 4^x=8\\
সমাধান: 4^x=8\\
or,\left(2^2\right)^x=2^3\\
or,2^{2x}=2^3\\
or,2x=3\\
or, x=\frac32
∴ নির্ণেয় সমাধান: x=\frac32
প্রশ্ন: ২০।। 2^{2x+1}=128\\
সমাধান:
2^{2x+1}=128\\
or,2^{2x}.2=128\\
or,2^{2x}=\frac{128}2\\
or,2^{2x}=64\\
or,2^{2x}=2^6\\
or,2x=6\\
or,x=\frac63\\
or,x=2
∴ নির্ণেয় সমাধান: x=2
প্রশ্ন: ২১।। \left(\sqrt3\right)^{x+1}=\left(\sqrt[3]3\right)^{2x-1}\\
সমাধান:
\left(\sqrt3\right)^{x+1}=\left(\sqrt[3]3\right)^{2x-1}\\
or,\left(3^\frac12\right)^{x+1}=\left(3^\frac13\right)^{2x-1}\\
or,3^\frac{x+1}2=3^\frac{2x-1}3\\
or,\frac{x+1}2=\frac{2x-1}3\\
or,4x-2=3x+3\\
or,4x-3x=3+2\\
or,x=5
∴ নির্ণেয় সমাধান: x=5
প্রশ্ন: 2^x+2^{1-x}=3\\or,2^x+2.2^{-x}=3\\
সমাধান:
2^x+2^{1-x}=3\\or,2^x+2.2^{-x}=3\\
or,2^x\left(2^x+2.2^{-x}\right)=3.2^x\\
or,2^{x+x}+2.2^0=3.2^x\\
or,2^{2x}+2=3.2^x\\
or,\left(2^x\right)^2-3.2^x+2=0\\
or,a^2-3a+2=0\\
or,a^2-2a-a+2=0\\
or,a(a-2)-1(a-2)=0\\
or,(a-2)(a-1)=0\\
or,a-2=0\\
or,a=2\\
or,2^x=2^1\\
or,x=1\\
nor,\;a-1=0\\
or,a=1\\
or,2^x=2^0\\
or,x=0
∴ নির্ণেয় সমাধান: x=0 অথবা, x=1
