How do you simplify \[4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}}\] and write it using only positive exponents?

Hint: We use the property of exponents that states that we can add the powers present on the numbers having the same bases. Here we have the same base ‘a’ and ‘b’, so we add the powers and calculate the value in power. Separate the constant values, values with base ‘a’ and with base ‘b’ and then solve.
* If a, m and n are any integer values then \[{a^m} \times {a^n} = {a^{m + n}}\]
* Any number having power 0 can be written as equal to 1 i.e. \[{a^0} = 1\]

Complete step by step solution:
We have to write the value of \[4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}}\]using a positive exponent
We know that when the bases are the same we can add the powers present in the numbers.
Collect the values with same variables
\[ \Rightarrow 4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}} = (4 \times 3)({a^3} \times {a^{ - 4}})({b^2} \times {b^{ - 3}})\]
Here we can write the value of the product by adding the powers .
\[ \Rightarrow 4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}} = (12)({a^{3 - 4}})({b^{2 - 3}})\]
Add the values in power
\[ \Rightarrow 4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}} = (12)({a^{ - 1}})({b^{ - 1}})\]
We know we can write \[{x^{ - 1}} = \dfrac{1}{x}\]
Then we can write right hand side of the equation using inverse
\[ \Rightarrow 4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}} = (12) \times \dfrac{1}{a} \times \dfrac{1}{b}\]
\[ \Rightarrow 4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}} = \dfrac{{12}}{{ab}}\]

\[\therefore \]The value of \[4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}}\]using a positive exponent is \[\dfrac{{12}}{{ab}}\].

Note: Many students make mistake of writing the final answer as \[12{a^{ - 1}}{b^{ - 1}}\], we know any value having negative power in the exponent can be written as value with same power but in the denominator. Keep in mind we are asked in the question to write the value using positive number in the exponent which means we have to write the value in exponent form, so we write the value \[{a^{ - 1}}\] and \[{b^{ - 1}}\] in form of fraction where we write ‘a’ and ‘b’ in the denominator.