Question

How do you simplify \[4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}}\]?

Answer 1

Hint: We use the concept that when the base is the same, powers can be collected and added. Collect the powers of ‘a’ and ‘b’ separately and solve the value in the power. Multiply the constant terms with constant terms.
* \[{p^m} \times {p^n} = {p^{m + n}}\], where the same base is p and m and n are the powers.

Complete step-by-step solution:
We have to simplify \[4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}}\]
We can clearly see that there are three terms in the value i.e. constant terms 4 and 3, variables ‘a’ and ‘b’ with powers.
We know all the values are in multiplication, so we can apply the property of collecting the powers of numbers having the same base. Collect the values that are constant in one bracket, values with the same base ‘a’ in one bracket and values with base ‘b’ in another bracket.
\[ \Rightarrow 4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}} = (4 \times 3).({a^3} \times {a^{ - 4}}).({b^2} \times {b^{ - 3}})\]
Now use the property of adding the powers when the bases are same in both the brackets
\[ \Rightarrow 4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}} = 12.({a^{3 + ( - 4)}}).({b^{2 + ( - 3)}})\]
Now we know the product of negative and positive term is always a negative term. Add the terms in the power for ‘a’ and ‘b’.
\[ \Rightarrow 4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}} = 12.({a^{3 - 4}}).({b^{2 - 3}})\]
Solve the value in power of ‘a’ and ‘b’.
\[ \Rightarrow 4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}} = 12.({a^{ - 1}}).({b^{ - 1}})\]
Multiply the terms together
\[ \Rightarrow 4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}} = 12{a^{ - 1}}{b^{ - 1}}\]
Now since power is same for both ‘a’ and ‘b’, we can collect the bases
\[ \Rightarrow 4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}} = 12{(ab)^{ - 1}}\]
Now we know any value having power as negative can be written in the denominator by making the power positive. So, for any ‘x’ we can write \[{x^{ - 1}} = \dfrac{1}{x}\]
\[ \Rightarrow 4{a^3}{b^2}.3{a^{ - 4}}{b^{ - 3}} = \dfrac{{12}}{{ab}}\]

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

Note:
Many students make the mistake of proceeding with the same pattern even when we have values with the same base but different powers in addition or subtraction. Keep in mind this rule only applies when the values having the same base are in multiplication, else this rule does not apply.