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**Hint:**Here, we have to evaluate the value of $\left( {\dfrac{1}{5}} \right)\log _2^{1024}$. We know that the value of \[\log _x^{{x^n}}\] is $n$. So, first of all we have to find the value $\log _2^{1024}$ by using the above given formula, then multiply the obtained value by $\dfrac{1}{5}$ to get the final result of the given problem.

**Complete step-by-step solution:**

Given that, evaluate the value of $\left( {\dfrac{1}{5}} \right)\log _2^{1024}$.

First of all find the value of $\log _2^{1024}$, so

Now, factorising $1024$, we can write

$1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^{10}}$.

We can write $\log _2^{1024}$as $\log _2^{{2^{10}}}$.

Now, applying the above given formula $\log _x^{{x^n}} = n$.

We can write $\log _2^{1024} = \log _2^{{2^{10}}} = 10$.

Now, we have to evaluate the value of $\left( {\dfrac{1}{5}} \right)\log _2^{1024}$.

So, $\left( {\dfrac{1}{5}} \right)\log _2^{1024} = \dfrac{1}{5} \times 10 = 2$

**Thus, the required value of $\left( {\dfrac{1}{5}} \right)\log _2^{1024}$ is $2$.**

**Note:**Some important formulae which are used to solve the problems related to logarithm are

(1) $\log _a^x + \log _a^y = \log _a^{xy}$

This formula is used to solve the problem like find the value of $\dfrac{1}{{\log _2^n}} + \dfrac{1}{{\log _3^n}} + \dfrac{1}{{\log _4^n}}$. For solving this problem first of all apply a basic logarithmic formula $\dfrac{1}{{\log _x^a}} = \log _a^x$ then by writing $\dfrac{1}{{\log _2^n}} = \log _n^2$ and similarly write other two terms and then apply this formula to get the answer.

(2) $\log _a^x - \log _a^y = \log _a^{\dfrac{x}{y}}$

This formula is used likewise formula (1) but only difference is in spite of multiplication do divisions.

(3) $\log _a^{\sqrt[n]{x}} = \dfrac{1}{n}\log _a^x$

This formula is used to solve the problem like finding the value of $\log _2^{\sqrt[{20}]{{1024}}}$. We have to solve this as similar to the above given procedure.

(4) $\log _a^x = \dfrac{{\log _c^x}}{{\log _c^a}}$.

Sometimes, we see that somewhere $\ln x$ and $\log x$ is given without base. “ $\ln x$” is called a natural logarithm whose base is $e$ that is $\ln x = \log _e^x$. If simply $\log x$ is given then it is commonly understood that its base is $10$.

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