Question

Express 729 as a power of 3.

Answer 1

Hint: Here, we need to express 729 as a power of 3. We will write the given number as a product of their factors. Then, we will simplify the product using rules of exponents, and express the number as a power of 3.

Formula Used:
We will use the rule of exponent, \[{a^b} \times {a^c} \times {a^d} = {a^{b + c + d}}\].

Complete step-by-step answer:
We will write the given number as a product of their factors, and then simplify the product to express it as a power of 3.
Let us use the divisibility tests of 2, 3, 5, etc. to find the factors of 729.
First, we will check the divisibility by 2.
We know that a number is divisible by 2 if it is an even number.
This means that any number that has one of the digits 2, 4, 6, 8, or 0 in the unit’s place, is divisible by 2.
We can observe that the number 729 has 9 at the unit’s place.
Therefore, the number 729 is not divisible by 2.
Next, we will check the divisibility by 3.
A number is divisible by 3 if the sum of its digits is divisible by 3.
We will add the digits of the number 729.
Thus, we get
\[7 + 2 + 9 = 18\]
Since the number 18 is divisible by 3, the number 729 is divisible by 3.
Dividing 729 by 3, we get
$\Rightarrow$ \[\dfrac{{729}}{3} = 243\]
Next, we will add the digits of the number 243.
Thus, we get
\[2 + 4 + 3 = 9\]
Since the number 9 is divisible by 3, the number 243 is divisible by 3.
Dividing 243 by 3, we get
$\Rightarrow$ \[\dfrac{{243}}{3} = 81\]
Next, we will add the digits of the number 81.
Thus, we get
\[8 + 1 = 9\]
Since the number 9 is divisible by 3, the number 81 is divisible by 3.
Dividing 81 by 3, we get
$\Rightarrow$ \[\dfrac{{81}}{3} = 27\]
We know that 27 is the product of 3 and 9.
Therefore, we can rewrite the number 729 as
\[729 = 3 \times 3 \times 3 \times 3 \times 9\]
Multiplying the pairs of 3 with each other, we get
\[\begin{array}{l} \Rightarrow 729 = 9 \times 9 \times 9\\ \Rightarrow 729 = {9^1} \times {9^1} \times {9^1}\end{array}\]
Rule of exponent: If two or more numbers with same base and different exponents are multiplied, the product can be written as \[{a^b} \times {a^c} \times {a^d} = {a^{b + c + d}}\].
Therefore, we can rewrite 729 as
\[\begin{array}{l} \Rightarrow 729 = {9^{1 + 1 + 1}}\\ \Rightarrow 729 = {9^3}\end{array}\]
\[\therefore \] We have expressed 729 as 9 raised to the power 3.

Note: We expressed 729 as 9 raised to the power 3. Here, 729 is called the cube of 9. When a number is raised to the power 3, the resulting product is its cube. Here, it is written power of 3, we might get confused by the statement and write 3 as a base and not as a power. If we write 3 as a base we get an answer as \[{3^5}\] which is wrong.