\(\log _{4}\left(\frac{1}{64}\right)\)
\(=\log _{4} 4^{-3}\)
\(=-3 \log _{4} 4\)
\(=-3\)

\(\log _{0.01} 0.000001\)
\(=\log _{0.01}(0.01)^{3}\)
\(=3 \cdot \log _{0.01}(0.01)\)
\(=3\)

\(\log _{\sqrt{6}} 216\)
\(=\log _{\sqrt{6}}(\sqrt{6})^{6}\)
\(=6 \log _{\sqrt{6}} \sqrt{6}\)
\(=6\)

\(\therefore\) \(1728\)
\(=2^{6} \times 3^{3}=2^{6} \times(\sqrt{3})^{6}\)
\(=(2 \sqrt{3})^{6}\)
\( \log _{2 \sqrt{3}} 1728=\log _{2 \sqrt{3}}(2 \sqrt{3})^{6} \)
\(=6 \cdot \log _{2 \sqrt{3}} 2 \sqrt{3}\left[\log _{a} a^{n}=n \cdot \log _{a}^{a}\right]\)
\(=6 \times 1\left[\because \log _{a} a=1\right]=6\)

মনে করি, \(625\)-এর লগারিদম \(4\) হলে নিধান \(n\) হবে।
\(\therefore \log _{n} 625=4\)
বা, \(n^{4}=625=(5)^{4} \)
বা, \(n=5\)

মনে করি, \(5832\)-র লগারিদম \(6\) হলে নিধান \(n\) হবে।

\(\therefore 5832=2^{3} \times 3^{6}\)
\(=(\sqrt{2})^{6} \times 3^{6}=(3 \sqrt{2})^{6}\)
\(\therefore \log _{n} 5832=6\)
বা, \(n^{6}=5832=(3 \sqrt{2})^{6}\)
\(\therefore n=3 \sqrt{2}\)

\(1+\log _{10} a=2 \log _{10} b\)
বা, \(\log _{10} 10+\log _{10} \mathrm{a}=\log _{10} \mathrm{~b}^{2}\left[\because 1=\log _{\mathrm{a}} \mathrm{a}\right]\)
বা, \(\log _{10}(10 a)=\log _{10} b^{2}\)
বা, \(10 a=b^{2}\)
বা, \(a=\frac{b^{2}}{10}\)
\(\therefore\) a কে b-এর দ্বারা প্রকাশ করে পাই \(a=\frac{b^{2}}{10}\)

\(3+\log _{10} x=2 \log _{10} y\)
বা, \(3 . \log _{10} 10+\log _{10} x=\log _{10} y^{2}\)
বা, \(\log _{10} 10^{3}+\log _{10} x=\log _{10} y^{2}\)
বা, \(\log _{10}\left(10^{3} x\right)=\log _{10} y^{2}\)
বা, \(1000 x=y^{2}\)
বা, \(x=\frac{y^{2}}{1000}\)
\(\therefore\) \(x\) কে y-এর দ্বারা প্রকাশ করে পাই \(x=\frac{y^{2}}{1000}\)

\( \log _{2}\left[\log _{2}\left\{\log _{3}\left(\log _{3} 27^{3}\right)\right\}\right] \)
\(=\log _{2}\left[\log _{2}\left\{\log _{3}\left(3 \log _{3} 3^{3}\right)\right\}\right] \)
\(=\log _{2}\left[\log _{2}\left\{\log _{3}\left(9 \log _{3} 3\right)\right\}\right] \)
\(=\log _{2}\left[\log _{2}\left\{\log _{3} 9\right\}\right]\quad {\left[\because \log _{3} 3=1\right]}\)
\(=\log _{2}\left[\log _{2}\left\{\log _{3} 3^{2}\right\}\right]=\log _{2}\left[\log _{2}\left\{2 \log _{3} 3\right\}\right]\)
\(=\log _{2}\left[\log _{2} 2\right] \quad {\left[\because \log _{3} 3=1\right]}\)
\(=\log _{2} 1\quad \left [\because \log _{2} 2=1\right]=0 \quad {\left[\because \log _{\mathrm{a}} 1=0\right]}\)

\(\frac{\log \sqrt{27}+\log 8-\log \sqrt{1000}}{\log 1.2}\)
\(=\frac{\log 3^{\frac{3}{2}}+\log 2^{3}-\log 10^{\frac{3}{2}}}{\log 1.2}\)
\(=\frac{\frac{3}{2} \log 3+3 \log 2-\frac{3}{2} \log 10}{\log 1.2}\)
\(=\frac{\frac{3}{2}(\log 3+2 \log 2-\log 10)}{\log 1.2}\)
\(=\frac{3}{2} \frac{(\log 3+\log 4-\log 10)}{\log 1.2}=\frac{3}{2} \frac{\log \left(\frac{3 \times 4}{10}\right)}{\log 1.2}\)
\(=\frac{3}{2} \frac{\log 1.2}{\log 1.2}=\frac{3}{2}\)

\(\log _{3} 4 \times \log _{4} 5 \times \log _{5} 6 \times \log _{6} 7 \times \log _{7} 3\)
\(=\log _{3} 5 \times \log _{5} 7 \times \log _{7} 3\left[\because \log _{a} b \times \log _{b} c=\log _{a} c\right]\)
\(=\log _{3} 7 \times \log _{7} 3=\log _{3} 3\)
\(=1\)
বিকল্প পদ্ধতি :
\(\log _{3} 4 \times \log _{4} 5 \times \log _{5} 6 \times \log _{6} 7 \times \log _{7} 3\)
\(=\frac{\log 4}{\log 3} \times \frac{\log 5}{\log 4} \times \frac{\log 6}{\log 5} \times \frac{\log 7}{\log 6} \times \frac{\log 3}{\log 7}\)
\(=1\)

\(\log _{10} \frac{384}{5}+\log _{10} \frac{81}{32}+3 \log _{10} \frac{5}{3}+\log _{10} \frac{1}{9}\)
\(=\log _{10} \frac{2^{7} \cdot 3}{5}+\log _{10} \frac{3^{4}}{2^{5}}+\log _{10} \frac{5^{3}}{3^{3}}+\log _{10} \frac{1}{3^{2}}\)
\(=\log _{10}\left(\frac{2^{7} \cdot 3}{5} \times \frac{3^{4}}{2^{5}} \times \frac{5^{3}}{3^{3}} \times \frac{1}{3^{2}}\right)\)
\(=\log _{10}\left(\frac{2^{7} \cdot 3^{5} \cdot 5^{3}}{2^{5} \cdot 3^{5} \cdot 5}\right)=\log _{10}\left(2^{2} \cdot 5^{2}\right)=\log _{10} 10^{2}\)
\(=2 \log _{10} 10=2\)

বামপক্ষ \(= \log \frac{75}{16}-2 \log \frac{5}{9}+\log \frac{32}{243}=\log 2\)
\(=\log \frac{3 \cdot 5^{2}}{2^{4}}-\log \frac{5^{2}}{3^{4}}+\log \frac{2^{5}}{3^{5}}=\log \left(\frac{3 \cdot 5^{2}}{2^{4}} \times \frac{3^{4}}{5^{2}} \times \frac{2^{5}}{3^{5}}\right)\)
\(=\left(\log \frac{2^{5} \cdot 3^{5} \cdot 5^{2}}{2^{4} \cdot 3^{5} \cdot 5^{2}}\right)\)
\(=\log 2\)
\(=\) ডানপক্ষ (প্রমানিত)

বামপক্ষ \(=\log _{10} 15\left(1+\log _{15} 30\right)+\frac{1}{2} \log _{10} 16\left(1+\log _{4} 7\right)\)
\(\quad \quad \quad -\log _{10} 6\left(\log _{6} 3+1+\log _{6} 7\right)\)
\(=\log _{10} 15+\log _{10} 15 \times \log _{15} 30+\log _{10} 16^{\frac{1}{2}} \)
\(\quad \quad \quad +\log _{10} 16^{\frac{1}{2}} \times \log _{4} 7-\log _{10} 6 \times \log _{6} 3-\log _{10} 6 \)
\(\quad \quad \quad \quad -\log _{6} 7 \times \log _{10} 6\)
\(=\log _{10} 15+\log _{10} 30+\log _{10} 4+\log _{10} 4 \times \log _{4} 7 \)
\(\quad \quad \quad -\log _{10} 3-\log _{10} 6-\log _{10} 7\)
\(=\log _{10}(5 \times 3)+\log _{10}(2 \times 3 \times 5)+\log _{10} 2^{2}\)
\(\quad \quad \quad +\log _{10} 7-\log _{10} 3-\log _{10}(3 \times 2)-\log _{10} 7\)
\( =\log _{10^{}} 5+\log _{10^{}} 3+\log _{10^{}}2+\log _{10^{}} 3+\log _{10^{}}5 \)
\( \quad \quad \quad \quad +2 \log _{10} 2-\log _{10} 3-\log _{10} 3-\log _{10} 2 \)
\( =2 \log _{10} 5+2 \log _{10} 2=\log _{10} 5^{2}+\log _{10^{}} 2^{2} \)
\(=\log _{10}\left(5^{2} \times 2^{2}\right)=\log _{10^{}} 10^{2}=2 \log _{10} 10\)
\(=2\)
\(=\) ডানপক্ষ(প্রমানিত)

বামপক্ষ \(=\log _{2} \log _{2} \log _{4} 256+2 \log _{\sqrt{2}} 2\)
\(=\log _{2} \log _{2} \log _{4} 4^{4}+2 \log _{\sqrt{2}}(\sqrt{2})^{2}\)
\(=\log _{2} \log _{2} 4 \cdot \log _{4} 4+2 \cdot 2 \log _{\sqrt{2}} \sqrt{2}\)
\(=\log _{2} \log _{2} 2^{2}+4\) \(\quad \)\( \left[\because \log _{a} a=1\right] \)
\(=\log _{2} 2 \cdot \log _{2} 2+4=\log _{2} 2+4 \)
\(=1+4=5\)
\(=\) ডানপক্ষ(প্রমানিত)

বামপক্ষ \(=\log _{x} 2 x \times \log _{y^{2}} y \times \log _{z^{2}} z\)
\(=\frac{1}{\log _{x} x^{2}} \times \frac{1}{\log _{y} y^{2}} \times \frac{1}{\log _{z} z^{2}}\)
\(=\frac{1}{2 \log _{x} x} \times \frac{1}{2 \log _{y} y} \times \frac{1}{2 \log _{z} z}\)
\(=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\)\(\quad\) \({\left[\because \log _{a} a=1\right]}\)
=\(\frac{1}{8}\)
= ডানপক্ষ(প্রমানিত)

বামপক্ষ \(=\log _{b^{3}} a \times \log _{c^{3}} b \times \log _{a^{3}} c\)
\(=\frac{1}{\log _{a} b^{3}} \times \frac{1}{\log _{b} c^{3}} \times \frac{1}{\log _{c} a^{3}}\)
\(=\frac{1}{3 \log _{a} b} \times \frac{1}{3 \log _{b} c} \times \frac{1}{3 \log _{c} a}\)
\(=\frac{1}{27} \log _{b} \mathrm{a} \times \log _{c} \mathrm{~b} \times \log _{a} \mathrm{c}\)
\(=\frac{1}{27} \log _{\mathrm{c}} \mathrm{a} \times \log _{\mathrm{a}} \mathrm{c} \quad\left[\because \log _{m} \mathrm{n} \times \log _{n} \mathrm{p}=\log _{m} \mathrm{p}\right]\)
\(=\frac{1}{27} \times 1\)
\(=\frac{1}{27}\)
= ডানপক্ষ(প্রমানিত)

বামপক্ষ \(=\frac{1}{\log _{x y}(x y z)}+\frac{1}{\log _{y z}(x y z)}+\frac{1}{\log _{z x}(x y z)}\)
\(=\log _{x y z} x y+\log _{x y z} y z+\log _{x y z} z x\)
\(=\log _{x y z}(x y \cdot y z \cdot z x)\)
\(=\log _{x y z}(x y z)^{2}\)
\(=2 \log _{x y z}(x y z)\)
\(=2.1\)
\(=2\)
=ডানপক্ষ(প্রমানিত)

বামপক্ষ \(=\log \frac{a^{2}}{b c}+\log \frac{b^{2}}{c a}+\log \frac{c^{2}}{a b}\)
\(=\log \left(\frac{a^{2}}{b c} \times \frac{b^{2}}{c a} \times \frac{c^{2}}{a b}\right)\)
\(=\log \left(\frac{a^{2} \times b^{2} \times c^{2}}{a^{2} \times b^{2} \times c^{2}}\right)\)
\(=\log 1=0\)
\(=\) ডানপক্ষ (প্রমানিত)

মনে করি,
\(x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}=p\)
বা, \(\log \left[x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}\right]=\log p\)
বা, \(\log x^{\log y-\log z}+\log y^{\log z-\log x}+\log z^{\log x-\log y}=\log p\)
বা, \((\log y-\log z) \log x+(\log z-\log x) \log y \)
\(\quad \quad \quad +(\log x-\log y) \log z=\log p\)
বা, \(\log x \cdot \log y-\log z \cdot \log x+\log y \cdot \log z-\log x \cdot \log y \)
\(\quad \quad \quad +\log x \cdot \log z-\log y \cdot \log z=\log p\)
বা, \(0=\log p \quad\)
বা, \( \log p=\log 1 \)
বা, \(p=1\)
\(\therefore x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}\)
\(=1\) (প্রমাণিত)

\(\log \frac{x+y}{5}=\frac{1}{2}(\log x+\log y)\)
বা, \(2 \log \frac{x+y}{5}=\log x+\log y\)
বা, \(\log \left(\frac{x+y}{5}\right)^{2}=\log (x y)\)
বা, \(\left(\frac{x+y}{5}\right)^{2}=x y\)
বা, \(\frac{x^{2}+y^{2}+2 x y}{25}=x y\)
বা, \(x^{2}+y^{2}+2 x y=25 x y\)
বা, \(x^{2}+y^{2}=23 x y\)
বা, \(\frac{x^{2}}{x y}+\frac{y^{2}}{x y}=23\)
বা, \(\frac{x}{y}+\frac{y}{x}=23 \) (প্রমাণিত)

\(a^{4}+b^{4}=14 a^{2} b^{2}\)
বা, \(\left(a^{2}\right)^{2}+\left(b^{2}\right)^{2}=14 a^{2} b^{2}\)
বা, \(\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}=14 a^{2} b^{2}\)
বা, \(\left(a^{2}+b^{2}\right)^{2}=16 a^{2} b^{2}\)
বা, \(\log \left(a^{2}+b^{2}\right)^{2}=\log (4 a b)^{2}\)
বা, \(2 \log \left(a^{2}+b^{2}\right)=2 \log (4 a b)\)
বা, \(\log \left(a^{2}+b^{2}\right)=\log 4+\log a+\log b\)
বা, \(\log \left(a^{2}+b^{2}\right)=\log a+\log b+\log 2^{2}\)
\(\therefore\) \(\log \left(a^{2}+b^{2}\right)=\log a+\log b+2 \log 2\) (প্রমাণিত)

\(\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}=\mathrm{k} \)(ধরি)
\(\therefore\) \(\log x=k(y-z), \log y=k(z-x) ; \log z=k(x-y)\)
\(\therefore\) \(\log x+\log y+\log z=k(y-z)+k(z-x)+k(x-y)\)
বা, \(\log (x y z)=k(y-z+z-x+x-y)\)
\(=0=\log 1\)
বা, \(x y z\)
\(=1 \) (প্রমাণিত)

\(\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k\) (ধরি)
\(\therefore\) \(\log x=k(b-c)\)
বা, \((b+c) \log x=k(b+c)(b-c)\)
বা, \(\log x^{(b+c)}=k\left(b^{2}-c^{2}\right)\ldots(i)\)
আবার, \(\log y=k(c-a)\)
বা, \((c+a) \log y=k(c+a)(c-a)\)
বা, \(\log y^{(c+a)}=k\left(c^{2}-a^{2}\right) \ldots(ii)\)
আবার, \(\log z=k(a-b)\)
বা, \((a+b) \log z=k(a+b)(a-b)\)
বা, \( \log z^{(a+b)}=k\left(a^{2}-b^{2}\right) \ldots(iii) \)
(i), (ii) ও (iii) যোগ করে পাই,
\(\therefore\) \(\log x^{b+c}+\log y^{c+a}+\log z^{a+b}\)
\(=k\left(b^{2}-c^{2}\right)+k\left(c^{2}-a^{2}\right)+k\left(a^{2}-b^{2}\right)\)
বা, \(\log \left(x^{b+c} \cdot y^{c+a} \cdot z^{c+a}\right)=0=\log 1\)
\(\therefore x^{b+c} \cdot y^{c+a} \cdot z^{a+b}=1\) (প্রমাণিত)
(b) \(\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k\) (ধরি)
\(\therefore\) \(\log x=k(b-c)\)
বা, \(\left(b^{2}+b c+c^{2}\right) \log x=k(b-c)\left(b^{2}+b c+c^{2}\right)\)
বা, \(\log x^{b^{2}+b c+c^{2}}=k\left(b^{3}-c^{3}\right) \ldots(i)\)
আবার, \(\frac{\log y}{c-a}=k\)
\(\therefore\) \(\log y=k(c-a)\)
বা, \(\left(c^{2}+c a+a^{2}\right) \log y=k(c-a)\left(c^{2}+c a+a^{2}\right)\)
বা, \( \log y^{c^{2}+c a+a^{2}}=k\left(c^{3}-a^{3}\right) \ldots(ii)\)
আবার, \(\frac{\log z}{a-b}=k\)
\(\therefore\) \(\log z=k(a-b)\)
বা, \(\left(a^{2}+a b+b^{2}\right) \log z=k(a-b)\left(a^{2}+a b+b^{2}\right)\)
বা, \(\log z^{a^{2}+a b+b^{2}}=k\left(a^{3}-b^{3}\right) \ldots(iii) \)
(i), (ii) ও (iii) যোগ করে পাই,
\(\log x^{b^{2}+b c+c^{2}}+\log y^{c^{2}+c a+a^{2}}+\log z^{a^{2}+a b+b^{2}}\)
\(=k\left(b^{3}-c^{3}+c^{3}-a^{3}+a^{3}-b^{3}\right)\)
বা, \(\log \left(x^{b^{2}+b c+c^{2}} \cdot y^{c^{2}+c a+a^{2}} \cdot z^{a^{2}+a b+b^{2}}\right)=\log 1\)
\(\therefore x^{b^{2}+b c+c^{2}} \cdot y^{c^{2}+c a+a^{2}} \cdot z^{a^{2}+a b+b^{2}}\)
\(=1\) (প্রমাণিত)

\(a^{3-x} \cdot b^{5 x}=a^{5+x} \cdot b^{3 x}\)
বা, \(\frac{b^{5 x}}{b^{3 x}}=\frac{a^{5+x}}{a^{3-x}}\)
বা, \(b^{5 x-3 x}=a^{5+x-3+x}\)
বা, \(b^{2 x}=a^{2 x+2}\)
বা, \(b^{2 x}=a^{2 x} \cdot a^{2}\)
বা, \(\left(\frac{b}{a}\right)^{2 x}=a^{2} \)
বা, \(\log \left(\frac{b}{a}\right)^{2 x}=\log a^{2}\)
বা, \(2 x \log \left(\frac{b}{a}\right)=2 \log a\)
\(\therefore\) \(x \log \left(\frac{b}{a}\right)=\log a \)(প্রমানিত)

\(\log _{8}\left[\log _{2}\left\{\log _{3}\left(4^{x}+17\right)\right\}\right]=\frac{1}{3}\)
বা, \(\log _{2}\left\{\log _{3}\left(4^{x}+17\right)\right\}=8^{\frac{1}{3}}\)\(\quad\) [\(\because \log _{a} n=p\) হলে \(a^{p}=n\) হয় ]
বা, \( \log _{2}\left\{\log _{3}\left(4^{x}+17\right)=2\right.\)
বা, \( \log _{3}\left(4^{x}+17\right)=2^{2}=4 \)
বা, \( 4^{x}+17=3^{4}=81 \quad \)
বা, \( 4^{x}=81-17=64=4^{3} \)
\(\therefore\) নির্ণেয় সমাধান \(x=3\)

\(\log _{8} x+\log _{4} x+\log _{2} x=11\)
বা, \(\frac{1}{\log _{x} 8}+\frac{1}{\log _{x} 4}+\log _{2} x=11\)
বা, \( \frac{1}{\log _{x} 2^{3}}+\frac{1}{\log _{x} 2^{2}}+\log _{2} x=11 \)
বা, \(\frac{1}{3 \log _{x^{2}} 2}+\frac{1}{2 \log _{x} 2}+\log _{2} x=11\)
বা, \(\frac{1}{3} \log _{2} x+\frac{1}{2} \log _{2} x+\log _{2} x=11\)
বা, \(\left(\frac{1}{3}+\frac{1}{2}+1\right) \log _{2} x=11\)
বা, \(\left(\frac{2+3+6}{6}\right) \log _{2} x=11\)
বা, \(\frac{11}{6} \log _{2} x=11\)
বা, \(\log _{2} x=11 \times \frac{6}{11}\)
বা, \(x=2^{6}=64\)
\(\therefore\) নির্ণেয় সমাধান \(x=64\)

\(\because 2^{3} < 10 < 2^{4}\)
বা, \(\log _{10^{}} 2^{3} < \log _{10} 10 < \log _{10^{2}} 2^{4}\)
বা, \( 3 \log _{10} 2<1<4 \log _{10} 2 \)
\(\therefore\) \(3 \log _{10^{}} 2 < 1\)
\(\therefore\) \(\log _{10^{}} 2 < \frac{1}{3} \ldots (i) \)
আবার, \(\because 1 < 4 \log _{10} 2\)
বা, \(\frac{1}{4} < \log _{10} 2\ldots(ii)\)
(i) ও (ii) হতে পাই, \(\frac{1}{4} < \log _{10} 2 < \frac{1}{3}\)(প্রমাণিত)

\(\log _{\sqrt{x}} 0.25=4\)
বা, \((\sqrt{x})^{4}=0.25 \quad\)
বা, \(x^{2}=(0.5)^{2}\)
\(\therefore x=0.5\)

\(\log _{10}(7 x-5)=2\)
বা, \((7 x-5)=10^{2} \)
বা, \(7 x=100+5=105\)
\(\therefore\) \(x=15\)

\(\log _{2} 3=a\)
বা, \(3 \log _{2} 3=3 a \)
বা, \(\log _{2} 3^{3}=3 a \)
বা, \(\log _{2} 27=3 a\)
বা, \(\frac{1}{\log _{27^{2}}}=3 a\)
বা, \(\frac{1}{3 \log _{27^{2}}}=a\)
বা, \(\frac{1}{\log _{27^{2}}{ }^{3}}=a\)
বা, \(\frac{1}{\log _{27} 8}=a\)
বা, \( \log _{8} 27=a\)

\(\log _{\sqrt{2}} x=a\)
বা, \(\frac{1}{\log _{x} \sqrt{2}}=a\)
বা, \(\frac{1}{3 \log _{x} \sqrt{2}}=\frac{a}{3}\)
বা, \(\frac{1}{\log _{x}(\sqrt{2})^{3}}=\frac{a}{3}\)
বা, \(\frac{1}{\log _{x} 2 \sqrt{2}}=\frac{a}{3} \)
বা, \(\log _{2 \sqrt{2}} x=\frac{a}{3}\)

\(\log _{x} \frac{1}{3}=-\frac{1}{3}\)
বা, \(x^{-\frac{1}{3}}=\frac{1}{3} \)
বা, \(\left(x^{-\frac{1}{3}}\right)^{-3}=\left(\frac{1}{3}\right)^{-3}\)
বা, \(x=3^{3}=27\)

\(\log _{4} \log _{4} \log _{4} 256\)
\(=\log _{4} \log _{4} \log _{4} 4^{4}=\log _{4} \log _{4} 4 \log _{4} 4\)
\(=\log _{4} \log _{4} 4\left[\because \log _{4} 4=1\right]\)
\(=\log _{4} 1=0\)

\(\log \frac{a^{n}}{b^{n}}+\log \frac{b^{n}}{c^{n}}+\log \frac{c^{n}}{a^{n}}\)
\(=\log \left(\frac{a^{n}}{b^{n}} \times \frac{b^{n}}{c^{n}} \times \frac{c^{n}}{a^{n}}\right)=\log 1=0\)

\(a^{\log _{a} x}=p\) (ধরি )
বা, \(\log _{a} a^{\log _{a} x}=\log _{a} p\)
বা, \(\log _{a} x \cdot \log _{a} a=\log _{a} p\)
বা, \(\log _{a} x=\log _{a} p \)
বা, \(x=p \)
বা, \(a^{\log _{a} x}=x \) (প্রমাণিত)

\(\log _{e} 2 \cdot \log _{x} 25=\log _{10} 16 \cdot \log _{e} 10\)
বা, \(\log _{e} 2 \log _{x} 25=\log _{e} 16\)
বা, \(\log _{x} 25=\frac{\log _{e} 16}{\log _{e} 2} \)
বা, \(\log _{x} 25=\log _{e} 16 \cdot \log _{2} e\)
বা, \(\log _{x} 25=\log _{2} 16\)
বা, \(\log _{x} 25=\log _{2} 2^{4}=4\)
বা, \(x^{4}=25\)
বা, \(x^{4}=(\sqrt{5})^{4}\)
বা, \(x=\sqrt{5}\)