Question

# 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 ${{\left( 1.25 \right)}^{12}}=\dfrac{{{5}^{12}}}{{{2}^{24}}}.$

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.