Question

How do you simplify \[\dfrac{{{x}^{2n}}\times {{x}^{3n-1}}}{{{x}^{n+2}}}\] leaving only positive exponents?

Answer 1

Hint: We have to remember \[{{x}^{n}}.{{x}^{m}}={{x}^{n+m}}.\] From this we can obtain \[{{x}^{n}}.{{x}^{-m}}={{x}^{n-m}}.\] Also, we have \[\dfrac{{{x}^{n}}}{{{x}^{m}}}={{x}^{n-m}}.\] And we can also write this as \[\dfrac{{{x}^{n}}}{{{x}^{m}}}=\dfrac{1}{{{x}^{m-n}}}.\] To simplify the above algebraic equation we may have to use all these identities together.

Complete step by step solution:
Consider the algebraic equation \[\dfrac{{{x}^{2n}}\times {{x}^{3n-1}}}{{{x}^{n+2}}}.\]
Let us see how to simplify this expression using the above given identities.
\[\Rightarrow \dfrac{{{x}^{2n}}\times {{x}^{3n-1}}}{{{x}^{n+2}}}=\dfrac{{{\left( {{x}^{n}} \right)}^{2}}{{x}^{3n-1}}}{{{x}^{n}}.{{x}^{2}}},\] we have used the identity \[{{\left( {{x}^{n}} \right)}^{m}}={{x}^{nm}}.\]
We can use cancellation of terms to cut off \[{{x}^{n}}\] from both the numerator and the denominator.
Let us recall how we eliminate the same terms up to the same power from the numerator and the denominator,
\[\Rightarrow \dfrac{{{x}^{2n}}\times {{x}^{3n-1}}}{{{x}^{n+2}}}=\dfrac{{{x}^{n}}.{{x}^{n}}.{{x}^{3n-1}}}{{{x}^{n}}.{{x}^{2}}}\]
After the cancellation we will get,
 \[\Rightarrow \dfrac{{{x}^{2n}}\times {{x}^{3n-1}}}{{{x}^{n+2}}}=\dfrac{{{x}^{n}}.{{x}^{3n-1}}}{{{x}^{2}}}\]
Now we have to use the identity \[{{x}^{n}}.{{x}^{m}}={{x}^{n+m}}\]
\[\Rightarrow \dfrac{{{x}^{2n}}\times {{x}^{3n-1}}}{{{x}^{n+2}}}=\dfrac{{{x}^{n+3n-1}}}{{{x}^{2}}}\]
In the next step we are using \[\dfrac{{{x}^{n}}}{{{x}^{m}}}={{x}^{n-m}}\]
\[\Rightarrow \dfrac{{{x}^{2n}}\times {{x}^{3n-1}}}{{{x}^{n+2}}}=\dfrac{{{x}^{n+3n-1-2}}}{1}\]
So, we will get
\[\Rightarrow \dfrac{{{x}^{2n}}\times {{x}^{3n-1}}}{{{x}^{n+2}}}={{x}^{4n-3}}\] which is the simplified form we are asked to find.
Now, there is another part in the question which asks us to write the simplified form of the given expression \[\dfrac{{{x}^{2n}}\times {{x}^{3n-1}}}{{{x}^{n+2}}}\] leaving only positive exponents.
In the simplified form we have obtained above, the exponent is \[4n-3.\]
The exponent \[4n-3\] is positive when,
\[\Rightarrow 4n-3>0\]
Now we are transposing 3 from the LHS of the symbol \[>\] to the RHS.
\[\Rightarrow 4n>3\]
Shifting \[4\]to the RHS from the LHS,
\[\Rightarrow n>\dfrac{3}{4}.\]
Thus, the exponent \[4n-3\] is positive when \[n>\dfrac{3}{4}.\]

Hence the simplified form of the given algebraic expression \[\dfrac{{{x}^{2n}}\times {{x}^{3n-1}}}{{{x}^{n+2}}}\] leaving only positive exponents is \[{{x}^{4n-3}}\] when \[n>\dfrac{3}{4}.\]

Note: We can simplify the given algebraic expression easily as follows:
In the first step we apply \[{{x}^{n}}{{x}^{m}}={{x}^{n+m}}\]
We get, \[\dfrac{{{x}^{2n}}\times {{x}^{3n-1}}}{{{x}^{n+2}}}=\dfrac{{{x}^{2n+3n-1}}}{{{x}^{2}}}\]
We apply \[\dfrac{{{x}^{n}}}{{{x}^{m}}}={{x}^{n-m}}\]
Then, \[\dfrac{{{x}^{2n}}\times {{x}^{3n-1}}}{{{x}^{n+2}}}={{x}^{2n+3n-1-\left( n+2 \right)}}\]
Consider the exponents only then,
\[\Rightarrow 2n+3n-1-\left( n+2 \right)=2n+3n-1-n-2\]
\[\Rightarrow 2n+3n-1-\left( n+2 \right)=4n-3\]
The exponent is positive only when \[4n-3>0\]
\[\Rightarrow 4n>3\]
\[\Rightarrow n>\dfrac{3}{4}.\]