Which is larger, \[{{\left( 2.5 \right)}^{6}}\] or \[{{\left( 1.25 \right)}^{12}}?\]

Hint: We have to compare both the terms to find which one is bigger. To do so, we will write them in the simplest form, that is, \[2.5=\dfrac{5}{2}\] and \[1.25=\dfrac{5}{4}.\] Then we can compare how \[{{\left( 2.5 \right)}^{6}}\] and \[{{\left( 1.25 \right)}^{12}}\] are related to each other by finding the value of \[{{\left( 2.5 \right)}^{6}}=\dfrac{{{5}^{6}}}{{{2}^{6}}}\] and \[{{\left( 1.25 \right)}^{12}}=\dfrac{{{5}^{12}}}{{{2}^{24}}}.\]

Complete step-by-step answer:
We are asked that which one is larger out of \[{{\left( 2.5 \right)}^{6}}\] and \[{{\left( 1.25 \right)}^{12}}.\] First, we will simplify both the terms.
\[2.5=\dfrac{25}{10}\]
On solving, we get,
\[\Rightarrow 2.5=\dfrac{5}{2}\]
So,
\[{{\left( 2.5 \right)}^{6}}={{\left( \dfrac{5}{2} \right)}^{6}}\]
We know that,
\[{{\left( \dfrac{a}{b} \right)}^{k}}=\dfrac{{{a}^{k}}}{{{b}^{k}}}\]
So, we can write,
\[\Rightarrow {{\left( 2.5 \right)}^{6}}=\dfrac{{{5}^{6}}}{{{2}^{6}}}.....\left( ii \right)\]
Similarly, we can write 1.25 as,
\[1.25=\dfrac{120}{10}\]
On solving, we will get,
\[\Rightarrow 1.25=\dfrac{5}{4}\]
So,
\[\Rightarrow {{\left( 1.25 \right)}^{12}}={{\left( \dfrac{5}{4} \right)}^{12}}\]
We know that,
\[{{\left( \dfrac{a}{b} \right)}^{k}}=\dfrac{{{a}^{k}}}{{{b}^{k}}}\]
\[\Rightarrow {{\left( 1.25 \right)}^{12}}=\dfrac{{{5}^{12}}}{{{4}^{12}}}\]
As we know that, \[{{2}^{2}}=4,\] we can write,
\[\Rightarrow {{\left( 1.25 \right)}^{12}}=\dfrac{{{5}^{12}}}{{{\left( {{2}^{2}} \right)}^{12}}}\]
We know that, \[{{\left( {{a}^{b}} \right)}^{c}}={{a}^{bc}}\]
\[\Rightarrow {{\left( 1.25 \right)}^{12}}=\dfrac{{{5}^{12}}}{{{2}^{24}}}\]
Now,
\[{{\left( 1.25 \right)}^{12}}=\dfrac{{{5}^{6+6}}}{{{2}^{6+18}}}\]
As, \[{{a}^{x+y}}={{a}^{x}}.{{a}^{y}},\] so
\[\Rightarrow {{\left( 1.25 \right)}^{12}}=\dfrac{{{5}^{6}}}{{{2}^{6}}}\times \dfrac{{{5}^{6}}}{{{2}^{18}}}......\left( ii \right)\]
Now, we know that,
\[{{5}^{2}}=25\]
\[{{2}^{5}}=2\times 2\times 2\times 2\times 2=32\]
Clearly, 25 < 32
\[\Rightarrow {{5}^{2}}<{{2}^{5}}\]
Simplifying further, we get,
\[\Rightarrow \dfrac{{{5}^{2}}}{{{2}^{5}}}<1\]
Now we know that if a < 1, then \[{{a}^{n}}\] is also less than 1 for all \[n\ge 1.\]
As \[\dfrac{{{5}^{2}}}{{{2}^{5}}}<1,\] so,
\[{{\left( \dfrac{{{5}^{2}}}{{{2}^{5}}} \right)}^{3}}<1\]
Simplifying using \[{{\left( {{a}^{b}} \right)}^{c}}={{a}^{bc}},\] we get,
\[\dfrac{{{5}^{6}}}{{{2}^{18}}}<1.....\left( iii \right)\]
Now, we use (iii) in (ii), we get,
\[\Rightarrow {{\left( 1.25 \right)}^{12}}=\dfrac{{{5}^{6}}}{{{2}^{6}}}\times \dfrac{{{5}^{6}}}{{{2}^{18}}}\]
As, \[\dfrac{{{5}^{6}}}{{{2}^{18}}}<1,\] so,
\[\Rightarrow {{\left( 1.25 \right)}^{12}}<\dfrac{{{5}^{6}}}{{{2}^{6}}}\times 1\]
\[\Rightarrow {{\left( 1.25 \right)}^{12}}<\dfrac{{{5}^{6}}}{{{2}^{6}}}\]
Using (i), we have, \[\dfrac{{{5}^{6}}}{{{2}^{6}}}={{\left( 2.5 \right)}^{6}}\]
So, we get,
\[{{\left( 1.25 \right)}^{12}}<{{\left( 2.5 \right)}^{6}}\]
So, we get, \[{{\left( 2.5 \right)}^{6}}\] is larger than \[{{\left( 1.25 \right)}^{12}}.\]

Note: We have to compare \[{{\left( 1.25 \right)}^{12}}\] with \[{{\left( 2.5 \right)}^{6}}.\] So, while expanding \[{{\left( 1.25 \right)}^{12}}=\dfrac{{{5}^{12}}}{{{2}^{24}}}\] into more simple term, we expand the term by keeping in the mind the value of \[{{\left( 2.5 \right)}^{6}}\] which is \[\dfrac{{{5}^{6}}}{{{2}^{6}}}.\] So, we separate this term out and then we will write as \[{{\left( 1.25 \right)}^{12}}=\dfrac{{{5}^{6}}}{{{2}^{6}}}.\dfrac{{{5}^{6}}}{{{5}^{18}}}.\] When we have this, we will check how other term affects the terms.