Question

How do you evaluate ${\left( { - 2} \right)^5}$?

Answer 1

Hint:
The given question is to simplify the given process and exponents. Powers and exponent are nothing but to solve the powers of the given exponent which is in the order of higher power for example ${4^2}$, the method to solve this type of question is that to multiply $4$ twice because the power $2$ means to multiply the base with itself 2 times. we get $4 \times 4 = 16$

Complete step by step solution:
The given question is to find out the value of given exponent. Power or exponent is nothing but the higher order power of some other number but the power can also be negative.
In the given question, we had to solve or simplify ${\left( { - 2} \right)^5}$ we had to make the question very clear that what is the different between ${\left( 2 \right)^5}$ and ${\left( { - 2} \right)^5}$ first one means we have to do the multiply $2$ five times and them use the negative sign in front of that which means ${\left( 2 \right)^5}$ means $2 \times 2 \times 2 \times 2 \times 2 = 32$, And ${\left( { - 2} \right)^5}$ means $\left( { - 32} \right)$ And in the second term, we have to solve ${\left( { - 2} \right)^5}$ Since negative sign is inside the bracket. Therefore $\left( { - 2} \right)$is taken as the whole-term is to be multiplied $5$ times.
Therefore ${\left( { - 2} \right)^5}$means $\left( { - 2} \right) \times \left( { - 2} \right) \times \left( { - 2} \right) \times \left( { - 2} \right) \times \left( { - 2} \right)$Since we had to multiply the number as well as negative sign. Therefore negative sign when multiplied by$5$times. So negative sign $4$times get cancel with each other and ${5^{th}}$sign is left also $2 \times 2 \times 2 \times 2 \times 2 = 32$

Therefore ${\left( { - 2} \right)^5} = - 32$

Note:
The given question was to simplify the given power and exponent therefore we had to use the formula which was required. Along with that some more formulas for solving power and exponents is
\[{a^m} \times {a^n} = {a^{mn}}{\text{ and }}\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\] Where $a$ , $b$,in and $c$ are any hyper of numbers either whole number, decimal or rational number.