Question

Find the value of ${\left( {625} \right)^{\dfrac{1}{4}}} + {\left( {25}\right)^{\dfrac{3}{2}}}$.

Answer 1

Hint: We can see this problem is from indices and powers. We need to simplify two different terms using the laws of exponents and powers. We will first express our base numbers in powers of five. Then, we will simplify the expression using the law of exponents and powers ${\left( {{a^x}} \right)^y} = \left( {{a^{xy}}} \right)$. Then, we will find the resultant expression by evaluating the final power of five.

Complete step by step solution:
We have, ${\left( {625} \right)^{\dfrac{1}{4}}} + {\left( {25} \right)^{\dfrac{3}{2}}}$.
So, we know the factorisation of $625$ is $625 = 5 \times 5 \times 5 \times 5$.
Also, the factorization of $25$ is $25 = 5 \times 5$
Expressing in exponential form, we get, $625 = {5^4}$ and $25 = {5^2}$.
So, we get the expression as,
${\left( {{5^4}} \right)^{\dfrac{1}{4}}} + {\left( {{5^2}} \right)^{\dfrac{3}{2}}}$
Using the law of exponents and powers ${\left( {{a^x}} \right)^y} = \left( {{a^{xy}}} \right)$, we get,
 \[ \Rightarrow \left( {{5^{4 \times \dfrac{1}{4}}}} \right) + \left( {{5^{2 \times \dfrac{3}{2}}}} \right)\]
Simplifying the powers of five, we get,
\[ \Rightarrow \left( {{5^1}} \right) + \left( {{5^3}} \right)\]
Calculating the powers of five and substituting the values, we get,
\[ \Rightarrow 5 + 125\]
\[ \Rightarrow 130\]
Therefore, the value of ${\left( {625} \right)^{\dfrac{1}{4}}} + {\left( {25} \right)^{\dfrac{3}{2}}}$ is \[130\].

Note:
These rules or laws of indices help us to minimize the problems and get the answer in very less time. These powers can be positive and negative but can be moulded according to our convenience while solving the problem. Also note that cube-root, square-root are fractions with \[1\] as numerator and respective root in denominator.