Q57 / 170 · Manipal (MET) 2021 · Mathematics

$\log x+\log x^3+\log x^5+\ldots .+\log x^{2 n-1}$ is equal to

  • A
    $2 n \log x$
  • B
    $(2 n-1) \log x$
  • Correct answer
    C
    $n^2 \log x$
  • D
    $\left(n^2+1\right) \log x$

Solution

$\log x+\log x^3+\log x^5+\ldots+\log x^{2 n-1}$ $\begin{aligned} & =\log \left(x \cdot x^3 \cdot x^5 \ldots x^{2 n-1}\right) \\ & =\log \left(x^{1+3+5+\ldots+2 n-1}\right) \\ & =\log x^{\frac{n}{2}[2+(n-1) 2]}=\log x^{n^2}=n^2 \log x\end{aligned}$

Hence, the correct answer is option: 'C'.

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