Question

How do you simplify the given term: $\dfrac{{{4}^{n}}}{{{3}^{n-1}}}$?

Answer 1

Hint: We start solving the problem by equating the given term to a variable. We then make use of the law of exponents that ${{a}^{n-m}}=\dfrac{{{a}^{m}}}{{{a}^{n}}}$, for $n > m$ and $a\ne 0$ to proceed through the problem. We then make the necessary calculations and then make use of the law of exponents that $\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}$, for $b\ne 0$ to get the required simplified form of the given term.

Complete step by step answer:
According to the problem, we are asked to simplify the given term: $\dfrac{{{4}^{n}}}{{{3}^{n-1}}}$.
Let us assume $d=\dfrac{{{4}^{n}}}{{{3}^{n-1}}}$ ---(1).
From the laws of exponents, we know that ${{a}^{n-m}}=\dfrac{{{a}^{m}}}{{{a}^{n}}}$, for $n > m$ and $a\ne 0$. Let us use this result in equation (1).
$\Rightarrow d=\dfrac{{{4}^{n}}\times 3}{{{3}^{n}}}$.
$\Rightarrow d=3\times \dfrac{{{4}^{n}}}{{{3}^{n}}}$ ---(2).
From the laws of exponents, we know that $\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}$, for $b\ne 0$. Let us use this result in equation (2).
$\Rightarrow d=3\times {{\left( \dfrac{4}{3} \right)}^{n}}$.
So, we have found the simplified form of the given term $\dfrac{{{4}^{n}}}{{{3}^{n-1}}}$ as $3\times {{\left( \dfrac{4}{3} \right)}^{n}}$.

$\therefore $ The simplified form of the given term $\dfrac{{{4}^{n}}}{{{3}^{n-1}}}$ is $3\times {{\left( \dfrac{4}{3} \right)}^{n}}$.

Note: Whenever we get this type of problems, we try to make use of the laws of exponents to get the required result. Here we have assumed that the value of n is greater than 1 to get the simplified form of the given term in the problem. We can also solve this problem by assuming that the value of n is less than 1 which may give the different answer. Similarly, we can expect problems to find the simplified form of the given term: $\dfrac{{{3}^{5}}\times {{20}^{6}}}{{{4}^{3}}\times {{5}^{5}}}$.