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# Which of the following is the value of $2\log x+2\log {{x}^{2}}+2\log {{x}^{3}}+...+2\log {{x}^{n}}$A. $\dfrac{n\left( n+1 \right)\log x}{2}$B. $n\left( n+1 \right)\log x$C. ${{n}^{2}}\log x$D. None of these

Last updated date: 09th Aug 2024
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Hint: To find the value of $2\log x+2\log {{x}^{2}}+2\log {{x}^{3}}+...+2\log {{x}^{n}}$ , first take 2 common outside and expand each log function using the formula $\log {{x}^{a}}=a\log x$ . From this, we will find $\log x$ to be common. Then, we will use the formula for series $1+2+3+...+n=\dfrac{n\left( n+1 \right)}{2}$ . After a few simplification steps, we will reach the correct option.

Complete step-by-step solution
We need to find the value of $2\log x+2\log {{x}^{2}}+2\log {{x}^{3}}+...+2\log {{x}^{n}}$ .
From the equation, we can observe that 2 is common to each term. So let us take 2 common from this. Hence, the above equation can be written as
$2\left( \log x+\log {{x}^{2}}+\log {{x}^{3}}+...+\log {{x}^{n}} \right)$
Let us now use the formula of logarithm.
We know that $\log {{x}^{a}}=a\log x$ . We can now write the above equation as
$2\left( \log x+2\log x+3\log x+...+n\log x \right)$
We can see that $\log x$ is a common term. Let us take it outside. Hence, the above equation becomes
$(1+2+3+...+n)2\log x$
Let us recollect that for a series $1+2+3+...+n=\dfrac{n\left( n+1 \right)}{2}$ . Now, let us apply this in the above equation. Thus, the above equation becomes
$\left( \dfrac{n\left( n+1 \right)}{2} \right)2\log x$
We can now cancel 2 from numerator and denominator. We will get
$n\left( n+1 \right)\log x$
Hence, the correct option is B.

Note: To solve this problem, the logarithmic rules must be known. One may make mistakes when writing the formula $\log {{x}^{a}}=a\log x$ as $\log {{x}^{a}}=\log ax$ . Also the general series equations must be known. You may make mistake when writing the formula $1+2+3+...+n=\dfrac{n\left( n+1 \right)}{2}$ as
$1+2+3+...+n=\dfrac{n\left( n-1 \right)}{2}$. One important point to keep in mind when solving these kinds of problems is that try to take the common terms outside from the beginning itself.