Question

Simplify the expression in t given below:
$\dfrac{25\times {{t}^{-4}}}{{{5}^{-2}}\times 10\times {{t}^{-8}}}\left( t\ne 0 \right)$

Answer 1

Hint: To solve the expression, we can use some basic properties of algebra and also, we can use some basic operations on multiplication and division. In algebra, we have a formula ${{t}^{-x}}=\dfrac{1}{{{t}^{x}}}$ where t and x can be any real or imaginary numbers.
Complete step-by-step answer:
Before proceeding with the question, we must know all the formulas that will be required to solve this question.
In algebra, we have a formula that can be used to simplify a number which is raised by a negative number. That formula is,
${{t}^{-x}}=\dfrac{1}{{{t}^{x}}}...........\left( 1 \right)$
If we have an expression that is written in the form of $\dfrac{x}{\left( \dfrac{y}{z} \right)}$ , this expression can be also written as,
\[\dfrac{x}{\left( \dfrac{y}{z} \right)}=\dfrac{xz}{y}...............\left( 2 \right)\]
If we have an expression that is written in the form of $\dfrac{\left( \dfrac{x}{y} \right)}{z}$ , then this expression can also be written as,
$\dfrac{\left( \dfrac{x}{y} \right)}{z}=\dfrac{x}{yz}................\left( 3 \right)$
Also, we have a formula $\dfrac{{{t}^{x}}}{{{t}^{y}}}={{t}^{x-y}}..................\left( 4 \right)$
In the question, we are required to simplify,
$\dfrac{25\times {{t}^{-4}}}{{{5}^{-2}}\times 10\times {{t}^{-8}}}$
Using equation $\left( 1 \right)$ , we can substitute ${{t}^{-4}}=\dfrac{1}{{{t}^{4}}},{{t}^{-8}}=\dfrac{1}{{{t}^{8}}}$ and ${{5}^{-2}}=\dfrac{1}{{{5}^{2}}}$. Substituting these in the above expression, we get,
$\dfrac{25\times \dfrac{1}{{{t}^{4}}}}{\dfrac{1}{{{5}^{2}}}\times 10\times \dfrac{1}{{{t}^{8}}}}$
Since ${{5}^{2}}=25$ , substituting ${{5}^{2}}=25$ in the above equation, we get,
$\dfrac{25\times \dfrac{1}{{{t}^{4}}}}{\dfrac{1}{25}\times 10\times \dfrac{1}{{{t}^{8}}}}$
Using equation $\left( 2 \right)$ and equation $\left( 3 \right)$ , the above expression can be simplified as,
$\begin{align}
  & \dfrac{25\times {{t}^{8}}\times 25}{10\times {{t}^{4}}} \\
 & \Rightarrow \dfrac{25\times 25}{10}\left( \dfrac{{{t}^{8}}}{{{t}^{4}}} \right) \\
\end{align}$
Using formula $\left( 4 \right)$ , we can write $\dfrac{{{t}^{8}}}{{{t}^{4}}}={{t}^{8-4}}={{t}^{4}}$ . Substituting this in the above expression, we get,
$\begin{align}
  & \dfrac{25\times 25}{10}{{t}^{4}} \\
 & \Rightarrow \dfrac{625}{10}{{t}^{4}} \\
 & \Rightarrow 62.5{{t}^{4}} \\
\end{align}$
Hence, the simplified form of the expression $\dfrac{25\times {{t}^{-4}}}{{{5}^{-2}}\times 10\times {{t}^{-8}}}$ is $62.5{{t}^{4}}$ .

Note: There is a possibility that one may commit a mistake while applying formula $\left( 2 \right)$ . There is a possibility that one may write it as \[\dfrac{x}{\left( \dfrac{y}{z} \right)}=\dfrac{xy}{z}\] instead of the correct formula \[\dfrac{x}{\left( \dfrac{y}{z} \right)}=\dfrac{xz}{y}\] . Similarly, there is a possibility that one may commit a mistake while applying the formula $\left( 3 \right)$ in the expression given in the question.