Simplify and express in exponential form: \[\left[ {{\left( -3 \right)}^{4}}\times {{\left( 1-3 \right)}^{2}} \right]\div {{\left( -3 \right)}^{7}}\]?

Hint: We start solving the problem by recalling the BODMAS order of performing operations which stands for Bracket, of, Division, Multiplication, Addition, and Subtraction. We then simplify the inner bracket having a combination of numbers with operations. We then remove the brackets for performing division and we make use of the result $ \dfrac{{{a}^{m}}}{{{a}^{n}}}=\dfrac{1}{{{a}^{n-m}}} $ if $ \left( n > m \right) $ to proceed through the problem. We then make the necessary arrangements to get the required answer.

Complete step by step answer:
According to the problem, we are asked to simplify the given \[\left[ {{\left( -3 \right)}^{4}}\times {{\left( 1-3 \right)}^{2}} \right]\div {{\left( -3 \right)}^{7}}\] and express it in exponential form.
Let us recall BODMAS order of performing operations. We know that BODMAS stands for Bracket, of, Division, Multiplication, Addition, and Subtraction.
So, this tells us that we need to perform operations present in innermost bracket first.
 $ \Rightarrow \left[ {{\left( -3 \right)}^{4}}\times {{\left( 1-3 \right)}^{2}} \right]\div {{\left( -3 \right)}^{7}}=\left[ {{\left( -3 \right)}^{4}}\times {{\left( -2 \right)}^{2}} \right]\div {{\left( -3 \right)}^{7}} $ .
Now, let us remove brackets (or parenthesis) to perform division operations.
\[\Rightarrow \left[ {{\left( -3 \right)}^{4}}\times {{\left( 1-3 \right)}^{2}} \right]\div {{\left( -3 \right)}^{7}}=\dfrac{{{\left( -3 \right)}^{4}}\times {{\left( -2 \right)}^{2}}}{{{\left( -3 \right)}^{7}}}\].
From the law of exponents, we know that $ \dfrac{{{a}^{m}}}{{{a}^{n}}}=\dfrac{1}{{{a}^{n-m}}} $ if $ \left( n > m \right) $ .
\[\Rightarrow \left[ {{\left( -3 \right)}^{4}}\times {{\left( 1-3 \right)}^{2}} \right]\div {{\left( -3 \right)}^{7}}=\dfrac{{{\left( -2 \right)}^{2}}}{{{\left( -3 \right)}^{7-4}}}\].
\[\Rightarrow \left[ {{\left( -3 \right)}^{4}}\times {{\left( 1-3 \right)}^{2}} \right]\div {{\left( -3 \right)}^{7}}=\dfrac{{{\left( -2 \right)}^{2}}}{{{\left( -3 \right)}^{3}}}\].
We know that $ {{\left( -a \right)}^{m}}={{\left( a \right)}^{m}} $ , if the value of m is even.
\[\Rightarrow \left[ {{\left( -3 \right)}^{4}}\times {{\left( 1-3 \right)}^{2}} \right]\div {{\left( -3 \right)}^{7}}=\dfrac{{{\left( 2 \right)}^{2}}}{{{\left( -3 \right)}^{3}}}\].
So, we have found the result of \[\left[ {{\left( -3 \right)}^{4}}\times {{\left( 1-3 \right)}^{2}} \right]\div {{\left( -3 \right)}^{7}}\] in exponential form as \[\dfrac{{{\left( 2 \right)}^{2}}}{{{\left( -3 \right)}^{3}}}\].
 $ \therefore $ The simplified exponential form of \[\left[ {{\left( -3 \right)}^{4}}\times {{\left( 1-3 \right)}^{2}} \right]\div {{\left( -3 \right)}^{7}}\] is \[\dfrac{{{\left( 2 \right)}^{2}}}{{{\left( -3 \right)}^{3}}}\].

Note:
 We should not express the obtained answer as $ \dfrac{-4}{27} $ which is not the exponential form, this should be kept in mind while solving this problem. Whenever we get the problem to express in exponential form, we should try to make use of the law of exponents. We should not confuse sign convention while solving this type of problem. Similarly, we can expect problems to express the result in logarithmic form.