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।। \[\sqrt{x^{-1}y}.\sqrt{y^{-1}z}.\sqrt{z^{-1}x} \]
\[\sqrt{x^{-1}y}.\sqrt{y^{-1}z}.\sqrt{z^{-1}x}\\\\=\sqrt{x^{-1}y.y^{-1}z.z^{-1}x}\\\\=\sqrt{x^{-1+1}.y^{1-1}.z^{1-1}}\\\\=\sqrt{x^0.y^0.z^0}\\\\=\sqrt{1.1.1}\\\\=\sqrt1\\\\=1\\\]
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
