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Evaluate the value of $\left( {\dfrac{1}{5}} \right)\log _2^{1024}$ .

Last updated date: 17th Jun 2024
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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 $10$.