Question

Solve the given expression, \[{{2}^{0}}+{{1}^{0}}-{{4}^{0}}-{{7}^{0}}\] .

Answer 1

Hint: We know the property that any number having 0 as its exponent is equal to 1. Here, we have the numbers 2, 1, 4, and 7 and each of these numbers has 0 as its exponents. Using this property we can simplify our given expression \[({{2}^{0}}+{{1}^{0}}-{{4}^{0}}-{{7}^{0}})\] .

Complete step-by-step answer:
According to the question, it is given that our expression is \[({{2}^{0}}+{{1}^{0}}-{{4}^{0}}-{{7}^{0}})\] . We have to solve this expression.
\[({{2}^{0}}+{{1}^{0}}-{{4}^{0}}-{{7}^{0}})\] ………………..(1)
In equation (1), we have the numbers 2, 1, 4, and 7 and each of these numbers has 0 as its exponents.
We know the property that any number having 0 as its exponent is equal to 1. That is,
\[{{x}^{0}}=1\]……………(2)
where x is any real number.
Here we have the numbers 2, 1, 4, and 7 and these all numbers are real numbers.
Replacing x by 2 in equation (2), we get
\[{{2}^{0}}=1\] ……………..(3)
Replacing x by 1 in equation (2), we get
\[{{1}^{0}}=1\] ……………..(4)
Replacing x by 4 in equation (2), we get
\[{{4}^{0}}=1\] ……………..(5)
Replacing x by 7 in equation (2), we get
\[{{7}^{0}}=1\] ……………..(6)
On adding equation (3), equation (4), equation (5), and equation (6), we get
\[{{2}^{0}}+{{1}^{0}}-{{4}^{0}}-{{7}^{0}}=1+1-1-1\]
\[\Rightarrow {{2}^{0}}+{{1}^{0}}-{{4}^{0}}-{{7}^{0}}=0\] ………………..(7)
We can see that the LHS of equation (7) is exactly equal to equation (1). So, the RHS of equation (7) should also be equal to equation (1).
Hence, the value of the expression \[\left( {{2}^{0}}+{{1}^{0}}-{{4}^{0}}-{{7}^{0}} \right)\] is 0.

Note: In this question, one can make a silly mistake in the property that any number having 0 as its exponent is equal to 0. Using this property one can replace the numbers \[{{2}^{0}}\] , \[{{1}^{0}}\] , \[{{4}^{0}}\] , and \[{{7}^{0}}\] by 0 in the expression \[\left( {{2}^{0}}+{{1}^{0}}-{{4}^{0}}-{{7}^{0}} \right)\] . This is wrong. Therefore, we have to keep the correct property in our mind.