Question

How do you write \[3 \times 3 \times 3 \times 3 \times y\]as an exponential form?

Answer 1

Hint: Exponent is the number of times it(number) is repeated. In order to solve this type of problem, you will need to separate the terms first and solve it individually. So when we take the series of 3s out to solve separately, we can conclude that $3 \times 3 \times 3 \times 3$ is equal to${3^4}$. And y equal to ${y^1}$. So, when we write them together we will get${3^4}{y^1}$, which is the exponential form.

Complete step-by-step solution:
The first step we have to do is to simplify the given form.
Here, there are two different terms in the question
One is 3 and other one is y
Now, a dot sign in a mathematical term like this means a product between two numbers.
Therefore, if we write the given form in a modified way, our new version of the question will be like:-
$3 \times 3 \times 3 \times 3 \times y$
Now, we know that repetitive multiplication of numbers with the same base can be written as the base number raised to a power. And that power is the number of times, it was multiplied. Basically, we will have to add the powers.
Therefore,
$3 \times 3 \times 3 \times 3$ can be written as ${3^4}$
So, if we replace \[3 \times 3 \times 3 \times 3\] by \[{3^4}\], we will get the modified form as:-
Now, similarly we can write y as\[{y^1}\], as it is multiplied only 1 times.
So, when we replace both the terms with the new version, we will get
\[3 \times 3 \times 3 \times 3 \times y = {3^4} \times {y^1}\] Which can also be written as:-
${3^4} \times {y^1} = {3^4}{y^1}$
And therefore, the exponential form of the given form \[3 \times 3 \times 3 \times 3 \times y\] will be ${3^4}{y^1}$.

Note: Addition of powers is done only when the numbers of the same base are multiplied with each other. In cases where two numbers of the same base are added, then we will add them separately. Mathematically, the first case is ${n^x} \times {n^y} = {n^{x + y}}$.