SSC General Math Exponent (সূচক) Chapter 4.1 Solution

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: $\left(\frac{x^a}{x^b}\right)^\frac1{ab}\times\left(\frac{x^b}{x^c}\right)^\frac1{bc}\times\left(\frac{x^c}{x^a}\right)^\frac1{ca}$

= $\left(\frac{x^a}{x^b}\right)^\frac1{ab}\times\left(\frac{x^b}{x^c}\right)^\frac1{bc}\times\left(\frac{x^c}{x^a}\right)^\frac1{ca}\\$

= $\left(x^{a-b}\right)^\frac1{ab}\times\left(x^{b-c}\right)^\frac1{bc}\times\left(x^{c-a}\right)^\frac1{ca}\\$

= $x^\frac{a-b}{ab}\times x^\frac{b-c}{bc}\times x^\frac{c-a}{ca}\\=x^\frac{ac-bc+ab-ac+bc-ab}{abc}\\$

= $x^\frac0{abc}\\$

= $x^0\\$

= 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: $\left(\frac{x^p}{x^q}\right)^{p+q-r}\times\left(\frac{x^q}{x^r}\right)^{q+r-p}\times\left(\frac{x^r}{x^p}\right)^{r+p-q}\\$

= $\left(x^{p-q}\right)^{p+q-r}\times\left(x^{q-r}\right)^{q+r-p}\times\left(x^{r-p}\right)^{r+p-q}\\$

= $x^{\left(p-q\right)\left(p+q-r\right)}\times x^{\left(q-r\right)\left(q+r-p\right)}\times x^{\left(r-p\right)\left(r+p-q\right)}\\$

= $x^{p^2+pq-pr-pq-q^2+qr}\times x^{q^2+qr-pq-qr-r^2+pr}\times x^{r^2+pr-qr-pr-p^2+pq}\\$

= $x^{{}^{p^2+pq-pr-pq-q^2+qr+q^2+qr-pq-qr-r^2+pr+r^2+pr-qr-pr-p^2+pq}}\\$

= $x^0\\$

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

Post Views: 2,462

SSC general math exponent (সূচক) chapter 4.1 solution