Question

Find the value of each of the following
 $(i){(16)^{\dfrac{1}{4}}}$
$(ii){(625)^{\dfrac{{ - 3}}{4}}}$

Answer 1

Hint: First, we have to simply solve the power which is placed over the given number.
After converting the given numbers $16,625$into the index form of $4,5$respectively, and then multiply it with the index placed over the given bracket of the expression $\dfrac{1}{4},\dfrac{{ - 3}}{4}$ respectively.
Since if the last number of the form ends with the number five, then the total number $625$will be divisible by the number five.
Properties of the exponent:
${({p^q})^r} = {({p^r})^q} = {p^r}^{ \times q}$
${p^a}.{p^b} = {p^{a + b}}$
${(\dfrac{p}{q})^r} = (\dfrac{{{p^r}}}{{{q^r}}})$

Complete step by step answer:
$(i){(16)^{\dfrac{1}{4}}}$
First, take the given expression and simplify the bracket part first using the index conversion of the square number.
When we reduce the square number (with the help of the perfect square, $16$can be reduced as the perfect square of $4$).
Now we can put the index power $2$as; ${(16)^{\dfrac{1}{4}}} \Rightarrow {({4^2})^{\dfrac{1}{4}}}$ (where $16 = 4 \times 4$)
Hence by the use of the property, ${({p^q})^r} = {({p^r})^q} = {p^r}^{ \times q}$
Using this property in the exponent, take the first and third parts of it we solve our expression as ${(16)^{\dfrac{1}{4}}} = {({4^2})^{\dfrac{1}{4}}} \Rightarrow {(4)^{2 \times }}^{\dfrac{1}{4}} = {(4)^{\dfrac{1}{2}}}$
Now converting the value by the use of division, we get ${(4)^{\dfrac{1}{2}}} = \sqrt 4 = 2$
Hence for $(i){(16)^{\dfrac{1}{4}}}$we get the value as $2$
$(ii){(625)^{\dfrac{{ - 3}}{4}}}$
Since the power is in the negative sign, to make it positive turn the numerator value to denominator we get, ${(625)^{\dfrac{{ - 3}}{4}}} = {(\dfrac{1}{{625}})^{\dfrac{3}{4}}}$
Again, the use of the property that, ${({p^q})^r} = {({p^r})^q} = {p^r}^{ \times q}$
We get, ${(\dfrac{1}{{625}})^{\dfrac{3}{4}}} = {[{(\dfrac{1}{{625}})^{\dfrac{1}{4}}}]^3}$
By the use of perfect square, we can convert the given as $625 = 5 \times 5 \times 5 \times 5 \Rightarrow {5^4}$
Substituting this we get, ${[{(\dfrac{1}{{625}})^{\dfrac{1}{4}}}]^3} = {[{(\dfrac{1}{{{5^4}}})^{\dfrac{1}{4}}}]^3}$then canceling the power four and power one by four, we get ${[{(\dfrac{1}{{{5^4}}})^{\dfrac{1}{4}}}]^3} = {[\dfrac{1}{5}]^3}$
Now by the property that ${(\dfrac{p}{q})^r} = (\dfrac{{{p^r}}}{{{q^r}}})$
Thus, we get, ${[\dfrac{1}{5}]^3} = \dfrac{1}{{125}}$
Therefore for $(ii){(625)^{\dfrac{{ - 3}}{4}}}$we get the value as $\dfrac{1}{{125}}$

Note: In layman terms, the index is the power raised to the given number and surd is the nth root of the given number $\sqrt[n]{b} = {(b)^{\dfrac{1}{n}}}$
The perfect square numbers, which is the numbers that obtain by multiplying any whole numbers (zero to infinity) twice, or the square of the given numbers yields a whole number like $\sqrt 4 = 2$or $4 = {2^2}$
Perfect square needs to satisfy that both the square of the given number and the root of the numbers are the same as above.