কষে দেখি 2
1. মান নির্নয় করোঃ
(i)\(\left(\sqrt[5]{8}\right)^\frac{5}{2}\times{(16)}^\frac{-3}{2}\)
সমাধানঃ
\(\left(\sqrt[5]{8}\right)^\frac{5}{2}\times{(16)}^\frac{-3}{2}\)
=\(\left(8^\frac{1}{5}\right)^\frac{5}{2}\times{(2^4)}^\frac{-3}{2}\)
=\(8^{\frac{1}{5}\times\frac{5}{2}}\times2^{4\times\left(-\frac{3}{2}\right)}\)
=\(\left(2^3\right)^\frac{1}{2}\times2^{-6}\)
=\(2^\frac{3}{2}\times2^{-6}\)
=\(2^{\frac{3}{2}-6}\)
=\(2^\frac{3-12}{2}\)
=\(2^{-\frac{9}{2}}\)
(ii) \(\left\{\left(125\right)^{-2}\times\left(16\right)^\frac{-3}{2}\right\}^\frac{-1}{6}\)
সমাধানঃ
=\(\left(8^\frac{1}{5}\right)^\frac{5}{2}\times{(2^4)}^\frac{-3}{2}\)
=\(8^{\frac{1}{5}\times\frac{5}{2}}\times2^{4\times\left(-\frac{3}{2}\right)}\)
=\(\left(2^3\right)^\frac{1}{2}\times2^{-6}\)
=\(2^\frac{3}{2}\times2^{-6}\)
=\(2^{\frac{3}{2}-6}\)
=\(2^\frac{3-12}{2}\)
=\(2^{-\frac{9}{2}}\)
(ii) \(\left\{\left(125\right)^{-2}\times\left(16\right)^\frac{-3}{2}\right\}^\frac{-1}{6}\)
সমাধানঃ
\(\left\{\left(125\right)^{-2}\times\left(16\right)^\frac{-3}{2}\right\}^\frac{-1}{6}\)
=\(\left\{\left(5^3\right)^{-2}\times\left(2^4\right)^\frac{-3}{2}\right\}^\frac{-1}{6}\)
=\(\left\{5^{-6}\times2^{-6}\right\}^\frac{-1}{6}\)
=\(\left\{\left(5\times2\right)^{-6}\right\}^\frac{-1}{6}\)
=\(\left(10\right)^{-6\times\left(\frac{-1}{6}\right)}\)
=10
(iii) \(4^\frac{1}{3}\times\left[2^\frac{1}{3}\times3^\frac{1}{2}\right]\div9^\frac{1}{4}\)
=\(\left\{\left(5^3\right)^{-2}\times\left(2^4\right)^\frac{-3}{2}\right\}^\frac{-1}{6}\)
=\(\left\{5^{-6}\times2^{-6}\right\}^\frac{-1}{6}\)
=\(\left\{\left(5\times2\right)^{-6}\right\}^\frac{-1}{6}\)
=\(\left(10\right)^{-6\times\left(\frac{-1}{6}\right)}\)
=10
(iii) \(4^\frac{1}{3}\times\left[2^\frac{1}{3}\times3^\frac{1}{2}\right]\div9^\frac{1}{4}\)
সমাধানঃ
\(4^\frac{1}{3}\times\left[2^\frac{1}{3}\times3^\frac{1}{2}\right]\div9^\frac{1}{4}\)
=\(\left(2^2\right)^\frac{1}{3}\times2^\frac{1}{3}\times3^\frac{1}{2}\div\left(3^2\right)^\frac{1}{4}\)
=\(2^\frac{2}{3}\times2^\frac{1}{3}\times3^\frac{1}{2}\div3^\frac{1}{2}\)
=\(2^{\frac{2}{3}+\frac{1}{3}}\times3^{\frac{1}{2}-\frac{1}{2}}\)
=\(2^\frac{3}{3}\times3^0\)
=2
2. সরল করোঃ
(i) \(\left(8a^3\div27x^{-3}\right)^\frac{2}{3}\times\left(64a^3\div27x^{-3}\right)^\frac{-2}{3}\)
\(4^\frac{1}{3}\times\left[2^\frac{1}{3}\times3^\frac{1}{2}\right]\div9^\frac{1}{4}\)
=\(\left(2^2\right)^\frac{1}{3}\times2^\frac{1}{3}\times3^\frac{1}{2}\div\left(3^2\right)^\frac{1}{4}\)
=\(2^\frac{2}{3}\times2^\frac{1}{3}\times3^\frac{1}{2}\div3^\frac{1}{2}\)
=\(2^{\frac{2}{3}+\frac{1}{3}}\times3^{\frac{1}{2}-\frac{1}{2}}\)
=\(2^\frac{3}{3}\times3^0\)
=2
2. সরল করোঃ
(i) \(\left(8a^3\div27x^{-3}\right)^\frac{2}{3}\times\left(64a^3\div27x^{-3}\right)^\frac{-2}{3}\)
সমাধানঃ
\(\left(8a^3\div27x^{-3}\right)^\frac{2}{3}\times\left(64a^3\div27x^{-3}\right)^\frac{-2}{3}\)
=\(\left\{\left(2a\right)^3\div\frac{27}{x^3}\right\}^\frac{2}{3}\times\left\{\left(4a\right)^3\div\frac{27}{x^3}\right\}^\frac{-2}{3}\)
=\(\left\{\left(2a\right)^3\times\left(\frac{x}{3}\right)^3\right\}^\frac{2}{3}\times\left\{\left(4a\right)^3\times\left(\frac{x}{3}\right)^3\right\}^\frac{-2}{3}\)
=\(\left\{\left(2a\times\frac{x}{3}\right)^3\right\}^\frac{2}{3}\times\left\{\left(4a\times\frac{x}{3}\right)^3\right\}^\frac{-2}{3}\)
=\(\left(2a\times\frac{x}{3}\right)^{3\times\frac{2}{3}}\times\left(4a\times\frac{x}{3}\right)^{3\times\frac{-2}{3}}\)
=\(\left(\frac{2ax}{3}\right)^2\times\left(\frac{4ax}{3}\right)^{-2}\)
=\(\left(\frac{2ax}{3}\right)^2\times\left(\frac{3}{4ax}\right)^2\)
=\(\left(\frac{2ax}{3}\times\frac{3}{4ax}\right)^2\)
=\(\left(\frac{1}{2}\right)^2= \frac{1}{4}\)
(ii) \(\left\{\left(x^{-5}\right)^\frac{2}{3}\right\}^\frac{-3}{\mathrm{10}}\)
\(\left(8a^3\div27x^{-3}\right)^\frac{2}{3}\times\left(64a^3\div27x^{-3}\right)^\frac{-2}{3}\)
=\(\left\{\left(2a\right)^3\div\frac{27}{x^3}\right\}^\frac{2}{3}\times\left\{\left(4a\right)^3\div\frac{27}{x^3}\right\}^\frac{-2}{3}\)
=\(\left\{\left(2a\right)^3\times\left(\frac{x}{3}\right)^3\right\}^\frac{2}{3}\times\left\{\left(4a\right)^3\times\left(\frac{x}{3}\right)^3\right\}^\frac{-2}{3}\)
=\(\left\{\left(2a\times\frac{x}{3}\right)^3\right\}^\frac{2}{3}\times\left\{\left(4a\times\frac{x}{3}\right)^3\right\}^\frac{-2}{3}\)
=\(\left(2a\times\frac{x}{3}\right)^{3\times\frac{2}{3}}\times\left(4a\times\frac{x}{3}\right)^{3\times\frac{-2}{3}}\)
=\(\left(\frac{2ax}{3}\right)^2\times\left(\frac{4ax}{3}\right)^{-2}\)
=\(\left(\frac{2ax}{3}\right)^2\times\left(\frac{3}{4ax}\right)^2\)
=\(\left(\frac{2ax}{3}\times\frac{3}{4ax}\right)^2\)
=\(\left(\frac{1}{2}\right)^2= \frac{1}{4}\)
(ii) \(\left\{\left(x^{-5}\right)^\frac{2}{3}\right\}^\frac{-3}{\mathrm{10}}\)
সমাধানঃ
\(\left\{\left(x^{-5}\right)^\frac{2}{3}\right\}^\frac{-3}{\mathrm{10}}\)
=\(\left\{x^{-5\times\frac{2}{3}}\right\}^\frac{-3}{\mathrm{10}}\)
=\(\left(x^\frac{-\mathrm{10}}{3}\right)^\frac{-3}{\mathrm{10}}\)
=\(\left(x\right)^{\left(\frac{-\mathrm{10}}{3}\right)\times\left(\frac{-3}{\mathrm{10}}\right)}\)
=\(x^1\)
\(\left\{\left(x^{-5}\right)^\frac{2}{3}\right\}^\frac{-3}{\mathrm{10}}\)
=\(\left\{x^{-5\times\frac{2}{3}}\right\}^\frac{-3}{\mathrm{10}}\)
=\(\left(x^\frac{-\mathrm{10}}{3}\right)^\frac{-3}{\mathrm{10}}\)
=\(\left(x\right)^{\left(\frac{-\mathrm{10}}{3}\right)\times\left(\frac{-3}{\mathrm{10}}\right)}\)
=\(x^1\)
= x
(iii) \(\left[\left\{\left(2^{-1}\right)^{-1}\right\}^{-1}\right]^{-1}\)
(iii) \(\left[\left\{\left(2^{-1}\right)^{-1}\right\}^{-1}\right]^{-1}\)
সমাধানঃ
\(\left[\left\{\left(2^{-1}\right)^{-1}\right\}^{-1}\right]^{-1}\)
=\(\left[\left\{\mathrm{2}^{(-1)\times(-1)}\right\}^{-1}\right]^{-1}\)
=\(\left[\left\{\mathrm{2}^1\right\}^{-1}\right]^{-1}\)
=\(\left[2^{1\times\left(-1\right)}\right]^{-1}\)
= \(2^{(-1)×(-1)}\)
= \(2^1\)
= 2
(v) \(\left(\frac{4^{m+\frac{1}{4}}\times\sqrt{2.2^m}}{2.\sqrt{2^{-m}}}\right)^\frac{1}{m}\)
\(\left[\left\{\left(2^{-1}\right)^{-1}\right\}^{-1}\right]^{-1}\)
=\(\left[\left\{\mathrm{2}^{(-1)\times(-1)}\right\}^{-1}\right]^{-1}\)
=\(\left[\left\{\mathrm{2}^1\right\}^{-1}\right]^{-1}\)
=\(\left[2^{1\times\left(-1\right)}\right]^{-1}\)
= \(2^{(-1)×(-1)}\)
= \(2^1\)
= 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
\(\left(\frac{4^{m+\frac{1}{4}}\times\sqrt{2.2^m}}{2.\sqrt{2^{-m}}}\right)^\frac{1}{m}\)
=\(\left(\frac{2^{2\left(m+\frac{1}{4}\right)}\times\left(2.2^m\right)^\frac{1}{2}}{2.2^\frac{-m}{2}}\right)^\frac{1}{m}\)
=\(\left(\frac{2^{2m+\frac{2}{4}}\times2^\frac{1}{2}.2^\frac{m}{2}}{2.2^\frac{-m}{2}}\right)^\frac{1}{m}\)
=\(\left(\frac{2^{2m+\frac{1}{2}+\frac{1}{2}+\frac{m}{2}}}{2.2^\frac{-m}{2}}\right)^\frac{1}{m}\)
=\(\left(2^{2m+1+\frac{m}{2}-1+\frac{m}{2}}\right)^\frac{1}{m}\)
=\(\left(2^{3m}\right)^\frac{1}{m}\)
=\(2^3\)
\(\left(\frac{4^{m+\frac{1}{4}}\times\sqrt{2.2^m}}{2.\sqrt{2^{-m}}}\right)^\frac{1}{m}\)
=\(\left(\frac{2^{2\left(m+\frac{1}{4}\right)}\times\left(2.2^m\right)^\frac{1}{2}}{2.2^\frac{-m}{2}}\right)^\frac{1}{m}\)
=\(\left(\frac{2^{2m+\frac{2}{4}}\times2^\frac{1}{2}.2^\frac{m}{2}}{2.2^\frac{-m}{2}}\right)^\frac{1}{m}\)
=\(\left(\frac{2^{2m+\frac{1}{2}+\frac{1}{2}+\frac{m}{2}}}{2.2^\frac{-m}{2}}\right)^\frac{1}{m}\)
=\(\left(2^{2m+1+\frac{m}{2}-1+\frac{m}{2}}\right)^\frac{1}{m}\)
=\(\left(2^{3m}\right)^\frac{1}{m}\)
=\(2^3\)
= 8
(vi) \(9^{-3}\times\frac{16^\frac{1}{4}}{6^{-2}}\times\left(\frac{1}{27}\right)^{-\frac{4}{3}}\)
(vi) \(9^{-3}\times\frac{16^\frac{1}{4}}{6^{-2}}\times\left(\frac{1}{27}\right)^{-\frac{4}{3}}\)
সমাধানঃ
\(9^{-3}\times\frac{16^\frac{1}{4}}{6^{-2}}\times\left(\frac{1}{27}\right)^{-\frac{4}{3}}\)
=\(\left(3\right)^{2\times\left(-3\right)}\times\frac{2^{4\times\frac{1}{4}}}{\left(2\times3\right)^{-2}}\times\left(\frac{1}{3}\right)^{3\times\left(-\frac{4}{3}\right)}\)
=\(3^{-6}\times\frac{2}{2^{-2}\times3^{-2}}\times\left(\frac{1}{3}\right)^{-4}\)
=\(\frac{1}{3^6}\times2\times2^2\times3^2\times3^4\)
=\(\frac{2^{1+2}\times3^{2+4}}{3^6}\)
=\(2^3\times3^{6-6}\)
=\(8×3^0\)
=8
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
\(9^{-3}\times\frac{16^\frac{1}{4}}{6^{-2}}\times\left(\frac{1}{27}\right)^{-\frac{4}{3}}\)
=\(\left(3\right)^{2\times\left(-3\right)}\times\frac{2^{4\times\frac{1}{4}}}{\left(2\times3\right)^{-2}}\times\left(\frac{1}{3}\right)^{3\times\left(-\frac{4}{3}\right)}\)
=\(3^{-6}\times\frac{2}{2^{-2}\times3^{-2}}\times\left(\frac{1}{3}\right)^{-4}\)
=\(\frac{1}{3^6}\times2\times2^2\times3^2\times3^4\)
=\(\frac{2^{1+2}\times3^{2+4}}{3^6}\)
=\(2^3\times3^{6-6}\)
=\(8×3^0\)
=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
\(\left(\frac{x^a}{x^b}\right)^{a^2+\mathrm{ab} +b^2}\times\left(\frac{x^b}{x^c}\right)^{b^2+bc+c^2}\times\left(\frac{x^c}{x^a}\right)^{c^2+ca+a^2}\)
=\(x^{(a-b)(a^2+ab+b^2)}\times x^{(b-c)(b^2+bc+c^2)}\times x^{(c-a)(c^2+ca+a^2)}\)
=\(x^{a^3-b^3}\times x^{b^3-c^3}\times x^{c^3-a^3}\)
=\(x^{a^3-b^3+b^3-c^3+\mathrm{.} \mathrm{c}^3-a^3}\)
=\(x^0\)
= 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
(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
∴ \(5^{24}<2^{60}<4^{36}<3^{48}\)
(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) \(\left(\frac{a^q}{a^r}\right)^p\times\left(\frac{a^r}{a^p}\right)^q\times\left(\frac{a^p}{a^q}\right)^r=1\)
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
(ii)\(\left(\frac{x^m}{x^n}\right)^{m+n}\times\left(\frac{x^n}{x^l}\right)^{n+l}\times\left(\frac{x^l}{x^m}\right)^{l+m}=1\)
\(\left(\frac{a^q}{a^r}\right)^p\times\left(\frac{a^r}{a^p}\right)^q\times\left(\frac{a^p}{a^q}\right)^r\)
=\(a^{\left(q-r\right)p}\times a^{\left(r-p\right)q}\times a^{\left(p-q\right)r}\)
=\(a^{pq-pr}\times a^{qr-pq}\times a^{pr-qr}\)
=\(a^{pq-pr+qr-pq+pr-qr}\)
=\(a^0\)
=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) \(\left(\frac{x^m}{x^n}\right)^{m+\mathrm{n-}l}\times\left(\frac{x^n}{x^l}\right)^{n+l-m}\times\left(\frac{x^l}{x^m}\right)^{l+\mathrm{m-n}}=1\)
\(\left(\frac{x^m}{x^n}\right)^{m+n}\times\left(\frac{x^n}{x^l}\right)^{n+l}\times\left(\frac{x^l}{x^m}\right)^{l+m}\)
=\(x^{(m-n)(m+n)}\times x^{(n-l)(n+l)}\times x^{(l-m)(l+m)}\)
=\(x^{m^2-n^2}\times x^{n^2-l^2}\times x^{l^2-m^2}\)
=\(x^{m^2-n^2+n^2-l^2+l^2-m^2}\)
=\(x^0\)
=1 [প্রমানিত]
(iii) \(\left(\frac{x^m}{x^n}\right)^{m+\mathrm{n-}l}\times\left(\frac{x^n}{x^l}\right)^{n+l-m}\times\left(\frac{x^l}{x^m}\right)^{l+\mathrm{m-n}}=1\)
সমাধানঃ
\(\left(\frac{x^m}{x^n}\right)^{m+n-l}\times\left(\frac{x^n}{x^l}\right)^{n+l-m}\times\left(\frac{x^l}{x^m}\right)^{l+m-n}\)
=\(x^{(m-n)(m+n-l)}\times x^{(n-l)(n+l-m)}\times x^{(l-m)(l+m-n)}\)
=\(x^{\left(m-n\right)\left(m+n\right)-\left(m-n\right)l}\times x^{\left(n-l\right)\left(n+l\right)-\left(n-l\right)m}\)
\(\times x^{(l-m)(l+m)-(l-m)n}\)
=\(x^{m^2-n^2-ml+nl}\times x^{n^2-l^2-mn+ml}\times x^{l^2-m^2-nl+mn}\)
=\(x^{m^2-n^2-ml+nl+n^2-l^2-mn+ml+l^2-m^2-nl+mn}\)
=\(x^0\)
=1 [প্রমানিত]
(iv) \(\left(a^\frac{1}{x-y}\right)^\frac{1}{x-z}\times\left(a^\frac{1}{y-z}\right)^\frac{1}{y-x}\times\left(a^\frac{1}{z-x}\right)^\frac{1}{z-y}=1\)
\(\left(\frac{x^m}{x^n}\right)^{m+n-l}\times\left(\frac{x^n}{x^l}\right)^{n+l-m}\times\left(\frac{x^l}{x^m}\right)^{l+m-n}\)
=\(x^{(m-n)(m+n-l)}\times x^{(n-l)(n+l-m)}\times x^{(l-m)(l+m-n)}\)
=\(x^{\left(m-n\right)\left(m+n\right)-\left(m-n\right)l}\times x^{\left(n-l\right)\left(n+l\right)-\left(n-l\right)m}\)
\(\times x^{(l-m)(l+m)-(l-m)n}\)
=\(x^{m^2-n^2-ml+nl}\times x^{n^2-l^2-mn+ml}\times x^{l^2-m^2-nl+mn}\)
=\(x^{m^2-n^2-ml+nl+n^2-l^2-mn+ml+l^2-m^2-nl+mn}\)
=\(x^0\)
=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]
\(\left(a^\frac{1}{x-y}\right)^\frac{1}{x-z}\times\left(a^\frac{1}{y-z}\right)^\frac{1}{y-x}\times\left(a^\frac{1}{z-x}\right)^\frac{1}{z-y}\)
=\(\left(a\right)^\frac{1}{(x-y)(x-z)}\times\left(a\right)^\frac{1}{(y-z)(y-x)}\times\left(a\right)^\frac{1}{(z-x)(z-y)}\)
=\(a^{\frac{1}{(x-y)(x-z)}+\frac{1}{(y-z)(y-x)}+\frac{1}{(z-x)(z-y)}}\)
=\(a^{\frac{1}{(x-y)(x-z)}-\frac{1}{(y-z)(x-y)}+\frac{1}{(x-z)(y-z)}}\)
=\(a^\frac{y-z+x-z+x-y}{(x-y)(x-z)(y-z)}\)
=\(a^\frac{0}{(x-y)(x-z)(y-z)}\)
=\(a^0\)
=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
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.\(a=xy^{p-1}, b=xy^{q-1}\) এবং \(c=xy^{r-1}\) হলে, দেখাই যে \(a^{q-r} b^{r-p} c^{p-q} = 1\)
=\(a^{y-z} b^{z-2y+z} c^{y-z}\)
=\(a^{y-z} b^{2(z-y)} c^{y-z}\)
=\(a^{y-z} (b^2)^{(z-y)} c^{y-z}\)
=\(a^{y-z} (ac)^{(z-y)} c^{y-z}\)
=\(a^{y-z} a^{z-y} c^{z-y} c^{y-z}\)
=\(a^{y-z+z-y} c^{z-y+y-z}\)
=\(a^0\times c^0\)
= 1
= ডানপক্ষ
\(\therefore\) বামপক্ষ = ডানপক্ষ [প্রমানিত]
6.\(a=xy^{p-1}, b=xy^{q-1}\) এবং \(c=xy^{r-1}\) হলে, দেখাই যে \(a^{q-r} b^{r-p} c^{p-q} = 1\)
সমাধানঃ
\(a=xy^{p-1}, b=xy^{q-1}\) এবং \(c=xy^{r-1}\)
\(a^{q-r} b^{r-p} c^{p-q}\)
=\((xy^{p-1})^{(q-r)} (xy^{q-1}){(r-p)} (xy^{r-1})^{(p-q )}\)
=\(x^{(q-r)}y^{(p-1)(q-r)} x^{(r-p)}y^{(q-1)(r-p)}x^{(p-q)}y^{(r-1)(p-q)}\)
=\(x^{q-r+r-p+p-q} y^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}\)
=\(x^0 y^{pq-pr-q+r+qr-pq-r+p+pr-qr-p+q}\)
=\(1.y^0\)
=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
\(a=xy^{p-1}, b=xy^{q-1}\) এবং \(c=xy^{r-1}\)
\(a^{q-r} b^{r-p} c^{p-q}\)
=\((xy^{p-1})^{(q-r)} (xy^{q-1}){(r-p)} (xy^{r-1})^{(p-q )}\)
=\(x^{(q-r)}y^{(p-1)(q-r)} x^{(r-p)}y^{(q-1)(r-p)}x^{(p-q)}y^{(r-1)(p-q)}\)
=\(x^{q-r+r-p+p-q} y^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}\)
=\(x^0 y^{pq-pr-q+r+qr-pq-r+p+pr-qr-p+q}\)
=\(1.y^0\)
=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 \(k\neq 0, 1, -1\)]
\(\therefore x^\frac{1}{a}=k\)
ধরি, \(x^\frac{1}{a}=y^\frac{1}{b}=z^\frac{1}{c}=k\) [যেখানে \(k\neq 0, 1, -1\)]
\(\therefore x^\frac{1}{a}=k\)
\(y^\frac{1}{b}=k\)
\(z^\frac{1}{c}=k\)
\(\therefore x=k^a, y=k^b, z=k^c\)
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
\(\therefore x=k^a, y=k^b, z=k^c\)
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 \(k\neq 0, 1, -1\)]
\(\therefore a^x=k\)
\(\therefore a^x=k\)
\(b^y=k\)
\(c^z=k\)
\(\therefore a=k^{\frac{1}{x}}, b=k^{\frac{1}{y}}, c=k^{\frac{1}{z}}\)
\(\therefore a=k^{\frac{1}{x}}, b=k^{\frac{1}{y}}, c=k^{\frac{1}{z}}\)
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
\((7^2)x=7^3\)
বা,\(7^{2x}=7^3\)
বা,2x=3
∴ x= \(\frac{3}{2}\)
(ii) \(2^{x+2}+2^{x-1}=9\)
(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,\(2^x = 2\)
\(2^{x+2}+2^{x-1}=9\)
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) \(2^{x+1}+2^{x+2}=48\)
∴ 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,\(2^x\times6=48\)
বা,\(2^x=\frac{48}{6}\)
বা,\(2^x =8\)
বা,\(2^x=2^3\)
বা,x=3
∴ নির্ণেয় সমাধান, x=3
(iv) \(2^{4x}.4^{3x-1}=\frac{4^{2x}}{2^{3x}}\)
\(2^{x+1}+2^{x+2}=48\)
বা,\(2^x.2^1+2^x.2^2=48\)
বা,\(2^x(2+4)=48\)
বা,\(2^x\times6=48\)
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) \(9\times81^x={27}^{2-x}\)
\(2^{4x}.4^{3x-1}=\frac{4^{2x}}{2^{3x}}\)
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) \(9\times81^x={27}^{2-x}\)
সমাধানঃ
\(9\times81^x={27}^{2-x}\)
বা,\(3^2\times\left(3^4\right)^x=\left(3^3\right)^{2-x}\)
বা,\(3^2\times3^{4x}=3^{6-3x}\)
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=\(\frac{4}{7}\)
\(9\times81^x={27}^{2-x}\)
বা,\(3^2\times\left(3^4\right)^x=\left(3^3\right)^{2-x}\)
বা,\(3^2\times3^{4x}=3^{6-3x}\)
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) \(6^{2x+4}=3^{3x}.2^{x+8}\)
\(2^{5x+4}+2^9=2^{10}\)
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) \(2^{\left(x^2-4\right)}=3^{\left(x^2-4\right)}\)
\(6^{2x+4}=3^{3x}.2^{x+8}\)
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) \(2^{\left(x^2-4\right)}=3^{\left(x^2-4\right)}\)
সমাধানঃ
\(2^{\left(x^2-4\right)}=3^{\left(x^2-4\right)}\)
বা,\(\frac{2^{\left(x^2-4\right)}}{3^{\left(x^2-4\right)}}=1\)
বা,\(\left(\frac{2}{3}\right)^{\left(x^2-4\right)}=\left(\frac{2}{3}\right)^0\)
বা, \(x^2-4 = 0\)
বা,\(x^2=4\)
∴ x=±2
∴ নির্ণেয় সমাধান, x=±2
10. বহু বিকল্পীয় প্রশ্ন(M.C.Q):
(i) \((0.243)^{0.2}×(10)^{0.6}\) এর মান
a. 0.3 b. 3 c. 0.9 d. 9
সমাধানঃ
\(2^{\left(x^2-4\right)}=3^{\left(x^2-4\right)}\)
বা,\(\frac{2^{\left(x^2-4\right)}}{3^{\left(x^2-4\right)}}=1\)
বা,\(\left(\frac{2}{3}\right)^{\left(x^2-4\right)}=\left(\frac{2}{3}\right)^0\)
বা, \(x^2-4 = 0\)
বা,\(x^2=4\)
∴ x=±2
∴ নির্ণেয় সমাধান, x=±2
10. বহু বিকল্পীয় প্রশ্ন(M.C.Q):
(i) \((0.243)^{0.2}×(10)^{0.6}\) এর মান
a. 0.3 b. 3 c. 0.9 d. 9
সমাধানঃ
\((0.243)^{0.2}×(10)^{0.6}\)
=\(\left(\frac{243}{1000}\right)^\frac{1}{5}×{(10)}^\frac{3}{5}\)
=\(\left(\frac{3^5}{{10}^3}\right)^\frac{1}{5}\times{(10)}^\frac{3}{5}\)
=\(\frac{3}{{10}^\frac{3}{5}}\times{10}^\frac{3}{5}\)
= 3
উত্তরঃ b. 3
(ii)\(2^\frac{1}{2}\times2^{-\frac{1}{2}}\times{(16)}^\frac{1}{2}\) এর মান
a. 1 b. 2 c. 4 d. \(frac{1}{2}\)
=\(\left(\frac{243}{1000}\right)^\frac{1}{5}×{(10)}^\frac{3}{5}\)
=\(\left(\frac{3^5}{{10}^3}\right)^\frac{1}{5}\times{(10)}^\frac{3}{5}\)
=\(\frac{3}{{10}^\frac{3}{5}}\times{10}^\frac{3}{5}\)
= 3
উত্তরঃ b. 3
(ii)\(2^\frac{1}{2}\times2^{-\frac{1}{2}}\times{(16)}^\frac{1}{2}\) এর মান
a. 1 b. 2 c. 4 d. \(frac{1}{2}\)
সমাধানঃ
\(2^\frac{1}{2}\times2^{-\frac{1}{2}}\times\left(16\right)^\frac{1}{2}\)
=\(2^{\frac{1}{2}-\frac{1}{2}}\times{(2^4)}^\frac{1}{2}\)
= \(2^0\times2^{4\times\frac{1}{2}}\)
= \(1×2^2\)
= 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
\(2^\frac{1}{2}\times2^{-\frac{1}{2}}\times\left(16\right)^\frac{1}{2}\)
=\(2^{\frac{1}{2}-\frac{1}{2}}\times{(2^4)}^\frac{1}{2}\)
= \(2^0\times2^{4\times\frac{1}{2}}\)
= \(1×2^2\)
= 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
\(4^x=8^3\)
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)\(4×5^x=500\) হলে, \(x^x\) এর মান
a. 8 b. 1 c. 64 d. 27
\(20^{-x}=\frac{1}{7}\)
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)\(4×5^x=500\) হলে, \(x^x\) এর মান
a. 8 b. 1 c. 64 d. 27
সমাধানঃ
\(4×5^x=500\)
বা,\(5^x=125\)
বা, \(5^x=5^3\)
বা,x=3
∴\(x^x=3^3=27\)
\(4×5^x=500\)
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।
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,\(4×5^5×0.01=5^x\)
বা,\(4×\frac{1}{100}\ =\frac{5^x}{5^5}\)
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) \(3×27^x=9^{x+4}\) হলে, x এর মান হিসাব করে লিখি।
\((5^5+0.01)^2-(5^5-0.01)^2=5^x\)
বা,\(4×5^5×0.01=5^x\)
বা,\(4×\frac{1}{100}\ =\frac{5^x}{5^5}\)
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) \(3×27^x=9^{x+4}\) হলে, x এর মান হিসাব করে লিখি।
সমাধানঃ
\(3×27^x=9^{x+4}\)
বা,\(3×(3^3)^x=(32)x+4\)
বা,\(3×3^{3x}=3^{2x+8}\)
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।
\(3×27^x=9^{x+4}\)
বা,\(3×(3^3)^x=(32)x+4\)
বা,\(3×3^{3x}=3^{2x+8}\)
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।
\(\sqrt[3]{\left(\frac{1}{64}\right)^\frac{1}{2}}\)
=\(\left(\frac{1}{64}\right)^{\frac{1}{2}\times\frac{1}{3}}\)
=\(\left(\frac{1}{2^6}\right)^\frac{1}{6}\)
=\(\left(\frac{1}{2}\right)^{6\times\frac{1}{6}}\)
=\(\frac{1}{2}\)
(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 \(3^{3^3}\)
\(3^{3^3}= 3^{27}\)
এবং \((3^3)^3=3^9\)
যেহেতু 27> 9
∴\(3^{3^3}>(3^3)^3\)
∴ বৃহত্তর সংখ্যাটি হল \(3^{3^3}\)
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