We need to compare
\({6^2}^{48},\,\,\,{25^2}^{47} = {5^{{2^1} \times {2^{47}}}} = {5^2}^{48},\,\)
\(\,{256^2}^{46} = {4^{4 \times {2^{46}}}} = {4^{{2^{48}}}} = {2^{{2^{49}}}},\)
\({6561^{{2^{45}}}} = {3^{8 \times {2^{45}}}} = {3^{{2^3} \times {2^{45}}}} = {3^{{2^{48}}}}\)
\(\,{65536^{{2^{44}}}} = {2^{16 \times {2^{44}}}} = {2^{{2^4} \times {2^{44}}}} = {2^{{2^{48}}}}\)

Let \({a^x} = {b^y} = {c^z} = k\), then
\(a = {k^{1/x}},\;\,b = {k^{1/y}},\,c = {k^{1/z}}\)
Given that \(abc = 1\), hence \({k^{1/x}}{k^{1/y}}{k^{1/z}} = 1 = {k^0}\)
\( \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0\) or \(xy + yz + zx = 0\)

Given that \({\left( {0.25} \right)^a} = {\left( {20} \right)^b} = {\left( 5 \right)^c} = k\)(say)
\( \Rightarrow \frac{1}{4} = {k^{1/a}},\,20 = {k^{1/b}},\,5 = {k^{1/c}}\)
Also \(\left( {\frac{5}{{20}}} \right) = \frac{1}{4} \Rightarrow \frac{{{k^{1/c}}}}{{{k^{1/b}}}} = {k^{1/a}}\)
\( \Rightarrow \frac{1}{c} - \frac{1}{b} = \frac{1}{a}\)

Multiplying and dividing by (17 – 1), then product can be simplified to:
\(\frac{{(17 - 1)(17 + 1)({{17}^2} + 1)....({{17}^{{2^{100}}}} + 1)}}{{(17 - 1)}}\)
\(\frac{{({{17}^{{2^{101}}}} - 1)}}{{(17 - 1)}} = \frac{{({{17}^{{2^{101}}}} - 1)}}{{16}}\)

Given that, \({4.7^x} = 1000\), hence
\(4.7 = {1000^{\frac{1}{x}}}\)
Similarly, \(0.047 = {1000^{\frac{1}{y}}}\)
Dividing the first equation by the second equation, we have
\(\frac{{4.7}}{{0.047}} = {1000^{\frac{1}{x} - \frac{1}{y}}}\)
\( \Rightarrow 100 = {1000^{\frac{1}{x} - \frac{1}{y}}}\)
\( \Rightarrow {10^2} = {10^{3\left( {\frac{1}{x} - \frac{1}{y}} \right)}}\)
Hence \(\frac{1}{x} - \frac{1}{y} = \frac{2}{3}\)

Assuming \({2^a} = x\), the given equation becomes
\({x^2}-12x + 32 = 0\) \( \Rightarrow x = 4,8\)
Thus \(a = 2,\,\,3\). There are 2 solutions.

Let \({x^3} = a\)\( \Rightarrow x = {a^{1/3}}\) …(1)
From the given relation \({x^a} = 3 \Rightarrow x = {3^{1/a}}\)
From the equation (1), \({a^{1/3}} = {3^{1/a}}\)
\( \Rightarrow {a^a} = {3^3}\) or \(a = 3\)and \(x = {3^{\frac{1}{3}}}\)

\({3^{324}} = {\left( {{3^4}} \right)^{81}}\)\( = {81^{81}}\)
Hence \(m = 81\) or \(\sqrt m = 9\)

The given inequality can be written as
\(0.25 \le {2^{n + 2}} \le 200\)
We know that \(0.25 = \frac{1}{4} = {2^{ - 2}}\) and \({2^8} = 256\)
\( \Rightarrow - 2 \le n + 2 \le 7\;{\rm{or}}\; - 4 \le n \le 5\)
Hence n can take 10 values.

Let us compare the first three numbers and take 12 powers of each so that the power of each becomes a natural number.
\({\left( {{2^{\frac{1}{2}}}} \right)^{12}},\,{\left( {{3^{\frac{1}{3}}}} \right)^{12}},\,{\left( {{5^{\frac{1}{4}}}} \right)^{12}}\)= \({2^6},\,{3^4},\,{5^3}\)
We see that \({5^3}\)is the biggest among the three numbers.
Now we can compare \({5^{\frac{1}{4}}}\;{\rm{and}}\;{6^{\frac{1}{5}}}\) using the same approach.

Given that x = \(\frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}\)= \(\frac{{{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{{5 - 3}}\)
= \(4 + \sqrt {15} \), similarly, y = \(4 - \sqrt {15} \)
Now \(xy = 1\) and \({x^2} + {y^2} = 62\)
\( \Rightarrow {x^3} + {y^3} = \left( {x + y} \right)\left( {{x^2} + {y^2}--xy} \right)\)
\( \Rightarrow {x^3} + {y^3} = \left( 8 \right)\left( {62--1} \right) = 488.\)

Multiplying numerator and denominator by \(\sqrt 2 \)
\( \Rightarrow \sqrt {52 - 2\sqrt {25 \times 27} } \Rightarrow \frac{{\sqrt {27} - \sqrt {25} }}{{10 - (\sqrt {75} + \sqrt 1 )}} = \frac{{3\sqrt 3 - 5}}{{9 - 5\sqrt 3 }}\)
Rationalizing, we get \(\frac{1}{{\sqrt 3 }}\).

The given surd can be written as
\(\left( {\frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 4\sqrt 3 - 4\sqrt 2 + 5\sqrt 2 }}} \right) = \frac{{3 + \sqrt 6 }}{{\sqrt {3 + \sqrt 2 } }} = \sqrt 3 \).

The number is \({10^{\frac{1}{5}}}{10^{\frac{2}{5}}}{......10^{\frac{n}{5}}} > 999\)
\( \Rightarrow {10^{\frac{1}{5}\left( {1 + 2 + 3 + ... + n} \right)}} > 999\)
\( \Rightarrow {10^{\frac{{n(n + 1)}}{{10}}}} > 999\)
Since 1000 > 999, hence \(\frac{{n(n + 1)}}{{10}} \ge 3\)
Or \(n(n + 1) \ge 30\) \( \Rightarrow n \ge 5\)

The simplified value is:
\({\log _{10}}{125^{1/3}} - {\log _{10}}{4^2} + {\log _{10}}32 - {\log _{10}}100\)
\( = {\log _{10}}\left( {\frac{{5 \times 32}}{{16 \times 100}}} \right) = {\log _{10}}\left( {\frac{1}{{10}}} \right) = - 1\)

\(\sqrt[m]{{\sqrt[n]{{\sqrt[n]{m}}}}} = \,n\)\( \Rightarrow \,{m^{\frac{1}{{m{n^2}}}}} = n \Rightarrow m = {n^{mn}}^2\)
\( \Rightarrow {\log _n}m = {\log _{_{^n}}}{n^{m{n^2}}} = m{n^{^{_2}}}\)

From the given relation,
\({\log _3}({\log _3}x) = {3^1} = 3\)
\( \Rightarrow {\log _3}x = {3^3} = 27\)
Hence \(x = {3^{{3^3}}} = {3^{27}} = {({3^3})^9} = {27^9}\)

\({\log _{30}}8 = 3{\log _{30}}2 = 3{\log _{30}}\left( {\frac{{30}}{{15}}} \right)\)
= \(3(1 - {\log _{30}}15) = 3(1 - {\log _{30}}3 - {\log _{30}}5)\)
= \(3(1--\alpha --\beta )\)

We know that \({\log _{{a^m}}}x = \frac{1}{m}{\log _a}x\)
Hence \({\log _{49}}343 = {\log _{{7^2}}}343 = \frac{1}{2}{\log _7}343 = \frac{3}{2}\)
Also \({\log _{343}}49 = \frac{1}{{{{\log }_{49}}343}} = \frac{2}{3}\)
Required answer = \(\frac{3}{2} - \frac{2}{3} = \frac{5}{6}\)

Value of the logarithm decreases when base increases for any value of a (> 1).

We can write \(5 = {\log _{10}}{10^5}\), hence the equation can be written as
\(\log \left( {\frac{{{{10}^5} \times {{\left( {\sqrt {1 - x} } \right)}^4}}}{{\sqrt {1 + x} }}} \right) = \log \left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)\)
\( \Rightarrow \log \left( {\frac{{{{10}^5} \times {{\left( {\sqrt {1 - x} } \right)}^4}}}{{\sqrt {1 + x} }}} \right) = \log \left( {\frac{1}{{\sqrt {(1 + x)(1 - x)} }}} \right)\)
\( \Rightarrow {10^5}{\left( {\sqrt {1 - x} } \right)^5} = 1\) or \(\sqrt {1 - x} = \frac{1}{{10}}\)
\( \Rightarrow x = \frac{{99}}{{100}}\;{\rm{or}}\;100x = 99\)

Given that \({\log _2}[3 + {\log _3}\{ 4 + {\log _4}(x - 1)\} ] = 2\)
\( \Rightarrow 3 + {\log _3}\{ 4 + {\log _4}(x - 1)\} = {2^2} = 4\)
\( \Rightarrow {\log _3}\{ 4 + {\log _4}(x - 1)\} = 1\)
\( \Rightarrow 4 + {\log _4}(x - 1) = 3\)
\( \Rightarrow {\log _4}(x - 1) = - 1\)
Hence \(x - 1 = \frac{1}{4}\) or \(4x = 5\)

The equation can be written as
\(\frac{{\log 5}}{{\log 4}} = \left( {\frac{{\log y}}{{\log 4}}} \right)\left( {\frac{{\log \sqrt 5 }}{{\log 6}}} \right)\)
\( \Rightarrow \frac{{\log 5}}{{\log 4}} = \left( {\frac{{\log y}}{{\log 4}}} \right)\left( {\frac{{\frac{1}{2}\log 5}}{{\log 6}}} \right)\)
\( \Rightarrow \frac{1}{2}\log y = \log 6\;{\rm{or}}\;y = 36\)

\({\log _{30}}45 = \frac{{{\rm{lo}}{{\rm{g}}_{\rm{2}}}45}}{{{{\log }_2}30}} = \frac{{2{{\log }_2}3 + {{\log }_2}5}}{{1 + {{\log }_2}3 + {{\log }_2}5}}\) …(1)
Since \({\log _2}5 = \frac{{{{\log }_3}5}}{{{{\log }_3}2}} = \frac{n}{{1/m}} = mn\)
Putting these values in (1), we get the answer.

\({\log _{ab}}x = \frac{1}{{{{\log }_x}ab}} = \frac{1}{{{{\log }_x}a + {{\log }_x}b}}\)
\( = \frac{1}{{\frac{1}{{{{\log }_a}x}} + \frac{1}{{{{\log }_b}x}}}}\)= \(\frac{{{{\log }_a}x{{\log }_b}x}}{{{{\log }_a}x + {{\log }_b}x}}\).

Let \(\frac{{\log \alpha }}{{\beta - \gamma }} = \frac{{\log \beta }}{{\gamma - \alpha }} = \frac{{\log \gamma }}{{\alpha - \beta }} = k\), then
\({\log _x}\alpha = k(\beta - \gamma ) \Rightarrow \alpha = {x^{k(\beta - \gamma )}}\)
Similarly, \(\beta = {x^{k(\gamma - \alpha )}}\), \(\gamma = {x^{k(\alpha - \beta )}}\)
Now \({\alpha ^\alpha }{\beta ^\beta }{\gamma ^\gamma } = {x^{k\alpha (\beta - \gamma ) + k\beta (\gamma - \alpha ) + k\gamma (\alpha - \beta )}} = {x^0} = 1\).

Suppose \(\frac{{x(y + z - x)}}{{\log x}} = \frac{{y(z + x - y)}}{{\log y}} = \frac{{z(x + y - z)}}{{\log z}} = a\)
\( \Rightarrow \log x = \frac{x}{a}(y + z - x),\log y = \frac{y}{a}(x + z - y)\)
and \(\log z = \frac{z}{a}(x + y - z)\)
Now \(y\log x + x\log y = \frac{{2xyz}}{a}\)
\(z\log y + y\log z = \frac{{2xyz}}{a}\)and \(z\log x + x\log z = \frac{{2xyz}}{a}\)
or \({x^y}{y^x} = {y^z}{z^y} = {x^z}{z^x}\)

The first equation can be written as
\({\log _{{2^2}}}{x^2} + {\log _4}y + {\log _4}z = 4\)
\( \Rightarrow {\log _4}({x^2}yz) = 4\;{\rm{or}}\;{x^2}yz = {4^4}\) ….. (1)
Similarly, second the third equations can be written as,
\({y^2}xz = {9^4}\) ….. (2)
and \({z^2}yx = {25^4}\) ….. (3)
Multiplying all the three equations, we have
\({x^4}{y^4}{z^4} = {(4 \times 9 \times 25)^4} \Rightarrow xyz = 900\) ….. (4)
Now using equation (2) and (4),
\(y = \frac{{{9^4}}}{{900}} = 7.29\)

The equation can be written as
\(\frac{1}{{{{\log }_{32}}a}} + \frac{1}{{{{\log }_{15}}a}} = 4\)
\( \Rightarrow {\log _a}32 + {\log _a}15 = 4\)
\( \Rightarrow {\log _a}(15 \times 32) = 4\)\( \Rightarrow {a^4} = 15 \times 32 = 480\)
Thus \(4 < a < 5\)

Taking log both sides
\({y^2}{\log _3}5\log 2 = {\log _2}3\log 5\)
\( \Rightarrow \frac{{{y^2}\log 5}}{{\log 3}} \times \log 2 = \frac{{\log 3}}{{\log 2}} \times \log 5\)
\( \Rightarrow {y^2} = {\left( {\frac{{\log 3}}{{\log 2}}} \right)^3}\)
\( \Rightarrow y = - \frac{{\log 3}}{{\log 2}} = - {\log _2}3 = {\log _2}\left( {\frac{1}{3}} \right)\)