Question

Which of the following is equal to 1?
(a) \[{{\left( 3 \right)}^{0}}+{{\left( 3 \right)}^{0}}\]
(b) \[{{\left( 3 \right)}^{0}}-{{\left( 3 \right)}^{0}}\]
(c) \[{{\left( 3 \right)}^{0}}\]
(d) \[{{\left( {{3}^{3}} \right)}^{0}}+{{\left( {{3}^{0}} \right)}^{3}}\]

Answer 1

Hint:Hint: To solve the question given above, we will check each option one by one. In checking each option, we will use certain exponential identities which are shown below:
\[{{\left( a \right)}^{b}}={{a}^{b}}\]
\[{{\left( {{a}^{b}} \right)}^{c}}={{a}^{b\times c}}\]
\[{{a}^{0}}=1\]
The option whose value evaluates to 1 will be the answer of this question.

Complete step-by-step answer:
In this question, we have to determine the option whose value will evaluate to 1. So, to solve this question, we will check each option by one. So, now we are going to check each option.
Option (a): \[{{\left( 3 \right)}^{0}}+{{\left( 3 \right)}^{0}}\]
To find the value of this expression, we will use certain exponential identities. First, we will remove the brackets of this term by the following identity:
\[{{\left( a \right)}^{b}}={{a}^{b}}\]
Thus, we get,
\[{{3}^{0}}+{{3}^{0}}\]
Now, we will use another identity which is as shown below:
\[{{a}^{0}}=1\]
Thus, we get,
= 1 + 1
= 2
Option (b): \[{{\left( 3 \right)}^{0}}-{{\left( 3 \right)}^{0}}\]
To find the value of this expression, first, we will remove the brackets from the term given in the option. For this, we will use the following identity.
\[{{\left( a \right)}^{b}}={{a}^{b}}\]
Thus, we get,
\[{{3}^{0}}-{{3}^{0}}\]
Now, we know that \[{{a}^{0}}=1.\] So, we will get,
= 1 – 1
= 0
Option (c): \[{{\left( 3 \right)}^{0}}\]
To find the value of this term, we will remove the bracket, thus we get,
\[={{3}^{0}}\]
Now, we will use the identity \[{{a}^{0}}=1.\] Thus, we get,
= 1
Option (d): \[{{\left( {{3}^{3}} \right)}^{0}}+{{\left( {{3}^{0}} \right)}^{3}}\]
To find the value of this term, we will use the following identity to remove the brackets:
\[{{\left( {{a}^{b}} \right)}^{c}}={{a}^{b\times c}}\]
Thus we get,
\[={{3}^{3\times 0}}+{{3}^{0\times 3}}\]
\[={{3}^{0}}+{{3}^{0}}\]
Now, we will use another identity, \[{{a}^{0}}=1.\] Thus, we get,

= 1 + 1
= 2
Hence, the option (c) is the right answer.

Note: The bracket removing identity which we have used in the options above is not valid everywhere. This identity is valid only when the base of the number is positive. If the base of the number is negative, we cannot apply this identity, i.e. \[{{\left( -a \right)}^{x}}\ne {{a}^{x}}\] if a < 0.