Question

By what number should $ {{3}^{-4}} $ be multiplied so that the product is 729? \[\]

Answer 1

Hint: We recall the definitions of exponent or power and law of exponent with same base $ \left( \dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} \right) $ . We convert the number 729 into an exponent of 3 using prime factorization. We assume the number of which should be multiplied width $ {{3}^{-4}} $ to get the product 729 as $ x $ and make an equation to solve for $ x $ .\[\]

Complete step by step answer:
We know from exponentiation that when any real number $ b $ is multiplied with itself say $ n $ times we can write as
\[b\times b\times ...\left( n\text{ times} \right)={{b}^{n}}\]
We read $ {{b}^{n}} $ as “ $ b $ to the power $ n. $ ” Here $ b $ is called base and $ n $ is called exponent, index, order or power.\[\]
We know from the law of exponent for quotients with the same base that while dividing exponential terms with the same base we subtract the exponent of denominator for the exponent of denominator. In symbols for some real numbers $ a,m,n $
\[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\]
 We are given the question the product has to be 729. Let us express 729 as an exponent of 3 using prime factorization. We have;
\[729=3\times 3\times 3\times 3\times 3\times 3={{3}^{6}}\]
Let us assume the number which we need to multiply $ {{3}^{-4}} $ to get the product 729 as $ x $ . So we have;
\[\begin{align}
  & x\times {{3}^{-4}}=729 \\
 & \Rightarrow x\times {{3}^{-4}}={{3}^{6}} \\
\end{align}\]
Let us divide both sides of the above step by $ {{3}^{-4}} $ and have;
\[\Rightarrow x=\dfrac{{{3}^{6}}}{{{3}^{-4}}}\]
We know use the law of exponent for quotient with same base $ \dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} $ in the above step for $ a=3,m=6,n=-4 $ to have;
\[\begin{align}
  & \Rightarrow x={{3}^{6-\left( -4 \right)}} \\
 & \Rightarrow x={{3}^{6+4}} \\
 & \Rightarrow x={{3}^{10}} \\
\end{align}\]
So $ {{3}^{10}} $ should be multiplied with $ {{3}^{-4}} $ so that the product is 729.\[\]

Note:
 We note that $ {{a}^{n}} $ is the reciprocal of $ {{a}^{-n}} $ and vice-versa. We can alternatively solve by multiplying the reciprocal of $ {{3}^{-4}} $ that is $ {{3}^{4}} $ both sides of the equation $ x\times {{3}^{-4}}={{3}^{6}} $ and then using the law of product with same base that is while multiplying exponential terms with same base we add the exponents $ \left( {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} \right) $ . We should be careful that base and exponent should ever be zero simultaneously. We should always remember divisibility rules to find the prime factorization of a number.