Question

Matrix Match Type

Column-1	column-2
1.Base of ${\left( { - 2} \right)^8}$	p) a
2.The value of ${\left( { - 1} \right)^{2010}} + {\left( { - 1} \right)^{2011}}$	q)${8^4}$
3.Fourth power of 8	r) 4096
4.Base of ${a^3}$	s) 0, t)-2

Answer 1

Hint: In ${x^y}$ where x is base and y is exponent corresponds to the number of times the base is used as a factor. In this question we can see that the base of ${\left( { - 2} \right)^8}$ is $ - 2$. Value of 1 having any power is equal to 1. Fourth power of 8 means ${8^4} = 8 \times 8 \times 8 \times 8$

Complete step-by-step answer:
Column-1
Base of ${\left( { - 2} \right)^8}$ is
We know that ${x^y}$ where x is base and y is exponent corresponds to the number of times the base is used as a factor.
Therefore, the base of ${\left( { - 2} \right)^8}$ is $ - 2$.
The value of ${\left( { - 1} \right)^{2010}} + {\left( { - 1} \right)^{2011}}$
We know that ${1^x} = 1$and even power of negative no. is even whereas odd power of negative no. is negative.
Therefore, ${\left( { - 1} \right)^{2010}} + {\left( { - 1} \right)^{2011}}$
$ \Rightarrow 1 + \left( { - 1} \right) = 0$
Fourth power of 8
We know that fourth power of 8 means ${8^4} = 8 \times 8 \times 8 \times 8$
Therefore, ${8^4} = 8 \times 8 \times 8 \times 8 = 4096$
Base of ${a^3}$
As similar to question no. 1 base = $a$
Answer- 1-t; 2-s; 3-q, r; 4-p

Note: In${x^y}$ where x is base and y is exponent corresponds to the number of times the base is used as a factor. Students should take care of the negative numbers. If the exponent is even for the negative number the value will be positive and for odd power value will be negative.