Question

Simplify the following and find the value of power: ${{\left( 25 \right)}^{7.5}}\times {{\left( 5 \right)}^{2.5}}\div {{\left( 125 \right)}^{1.5}}={{5}^{?}}$
A. $8.5$
B. $13$
C. $16$
D. $17.5$

Answer 1

Hint: At first, we apply the BODMAS rule and solve ${{\left( 5 \right)}^{2.5}}\div {{\left( 125 \right)}^{1.5}}$ . $125$ can be written as ${{5}^{3}}$ so, expression becomes $\dfrac{{{\left( 5 \right)}^{2.5}}}{{{\left( {{\left( 5 \right)}^{3}} \right)}^{1.5}}}$ . This becomes $\dfrac{{{\left( 5 \right)}^{2.5}}}{{{\left( 5 \right)}^{4.5}}}$ according to ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . We then write it as ${{\left( 5 \right)}^{2.5-4.5}}={{\left( 5 \right)}^{-2}}$ according to $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ . Then, we do ${{\left( 25 \right)}^{7.5}}\times {{\left( 5 \right)}^{-2}}$ . $25$ can be written as ${{5}^{2}}$ so, expression becomes ${{\left( {{\left( 5 \right)}^{2}} \right)}^{7.5}}\times {{\left( 5 \right)}^{-2}}$ , which can be written as ${{\left( 5 \right)}^{15}}\times {{\left( 5 \right)}^{-2}}$ according to ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . Now, we can write it as ${{\left( 5 \right)}^{15-2}}$ according to ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ . Simplifying this gives the answer.

Complete step by step answer:
In this problem, we are given the mathematical expression,
${{\left( 25 \right)}^{7.5}}\times {{\left( 5 \right)}^{2.5}}\div {{\left( 125 \right)}^{1.5}}$
The above expression involves multiplication and division. So, we have to use the BODMAS (Bracket Of Division Multiplication Addition Subtraction) rule which gives the preference of operations in decreasing order. So, we have to solve ${{\left( 5 \right)}^{2.5}}\div {{\left( 125 \right)}^{1.5}}$ first. This can be written as,
$\Rightarrow {{\left( 5 \right)}^{2.5}}\div {{\left( 125 \right)}^{1.5}}=\dfrac{{{\left( 5 \right)}^{2.5}}}{{{\left( 125 \right)}^{1.5}}}$
Now, $125$ can be written as ${{5}^{3}}$ . So, the expression becomes,
$\Rightarrow \dfrac{{{\left( 5 \right)}^{2.5}}}{{{\left( 125 \right)}^{1.5}}}=\dfrac{{{\left( 5 \right)}^{2.5}}}{{{\left( {{\left( 5 \right)}^{3}} \right)}^{1.5}}}$
Now, we know that ${{\left( {{a}^{m}} \right)}^{n}}$ can be written as ${{a}^{mn}}$ . So, the expression becomes,
$\Rightarrow \dfrac{{{\left( 5 \right)}^{2.5}}}{{{\left( {{\left( 5 \right)}^{3}} \right)}^{1.5}}}=\dfrac{{{\left( 5 \right)}^{2.5}}}{{{\left( 5 \right)}^{4.5}}}$
Now, we know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}$ can be written as ${{a}^{m-n}}$ . So, the expression becomes,
$\Rightarrow \dfrac{{{\left( 5 \right)}^{2.5}}}{{{\left( 5 \right)}^{4.5}}}={{\left( 5 \right)}^{2.5-4.5}}={{\left( 5 \right)}^{-2}}$
The given expression thus becomes,
${{\left( 25 \right)}^{7.5}}\times {{\left( 5 \right)}^{-2}}$
Now, $25$ can be written as ${{5}^{2}}$ . So, the above expression becomes,
$\Rightarrow {{\left( 25 \right)}^{7.5}}\times {{\left( 5 \right)}^{-2}}={{\left( {{\left( 5 \right)}^{2}} \right)}^{7.5}}\times {{\left( 5 \right)}^{-2}}$
Now, we know that ${{\left( {{a}^{m}} \right)}^{n}}$ can be written as ${{a}^{mn}}$ . So, the expression becomes,
$\Rightarrow {{\left( {{\left( 5 \right)}^{2}} \right)}^{7.5}}\times {{\left( 5 \right)}^{-2}}={{\left( 5 \right)}^{15}}\times {{\left( 5 \right)}^{-2}}$
Now, we know that ${{a}^{m}}\times {{a}^{n}}$ can be written as ${{a}^{m+n}}$ . So, the expression becomes,
$\Rightarrow {{\left( 5 \right)}^{15}}\times {{\left( 5 \right)}^{-2}}={{\left( 5 \right)}^{15-2}}={{\left( 5 \right)}^{13}}$
Therefore, we can conclude that the answer to the problem is $13$.

So, the correct answer is “Option B”.

Note: We can also solve the problem in another way. In this process, we simply use the calculator to solve. We first find the values of ${{\left( 25 \right)}^{7.5}},{{\left( 5 \right)}^{2.5}},{{\left( 125 \right)}^{1.5}}$ using calculator, which come out as $3.0517\times {{10}^{10}}$ , $55.9016$ and $1397.5424$ respectively. Then, we evaluate ${{\left( 5 \right)}^{2.5}}\div {{\left( 125 \right)}^{1.5}}$ which is $\dfrac{55.9016}{1397.5424}=0.04$ . Then, we find the value of ${{\left( 25 \right)}^{7.5}}\times 0.04$ which is $3.0517\times {{10}^{10}}\times 0.04=1220703125$ . Now, we need to prime factorise $1220703125$ . It gives,
$\begin{align}
  & 5\left| \!{\underline {\,
  1220703125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  244140625 \,}} \right. \\
 & 5\left| \!{\underline {\,
  48828125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  9765625 \,}} \right. \\
 & 5\left| \!{\underline {\,
  1953125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  390625 \,}} \right. \\
 & 5\left| \!{\underline {\,
  78125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  15625 \,}} \right. \\
 & 5\left| \!{\underline {\,
  3125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  625 \,}} \right. \\
 & 5\left| \!{\underline {\,
  125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  25 \,}} \right. \\
 & ~~~\left| \!{\underline {\,
  5 \,}} \right. \\
\end{align}$
Which is ${{5}^{13}}$ . So, $13$ is the answer.