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If $x=1+\frac{1}{2 \times 1 !}+\frac{1}{4 \times 2 !}+\frac{1}{8 \times 3 !}+\ldots$ and
$y=1+\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}+\frac{x^{6}}{3 !}+\ldots$
Then, the value of loge $y$ is
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$y=1+\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}+\frac{x^{6}}{3 !}+\ldots$
Then, the value of loge $y$ is
Solution:
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Verified Answer
The correct answer is:
$e$
Given,
$x=1+\frac{1}{2 \times 1 !}+\frac{1}{4 \times 2 !}+\frac{1}{8 \times 3 !}+\ldots$
$\Rightarrow \quad x=1+\frac{(1 / 2)^{\prime}}{1 !}+\frac{(1 / 2)^{2}}{2 !}+\frac{(1 / 2)^{3}}{3 !}+\ldots$
$\Rightarrow \quad x=e^{1 / 2} \Rightarrow x^{2}=e$
and $\quad y=1+\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}+\frac{x^{6}}{3 !}+\ldots$
$\Rightarrow \quad y=1+\frac{\left(x^{2}\right)^{1}}{1 !}+\frac{\left(x^{2}\right)^{2}}{2 !}+\frac{\left(x^{2}\right)^{3}}{3 !}+\ldots$
$\Rightarrow \quad y=e^{x^{2}}=e^{e} \quad$ [from Eq. (i)]
Taking log on both sides, we get
log$y=e$
$x=1+\frac{1}{2 \times 1 !}+\frac{1}{4 \times 2 !}+\frac{1}{8 \times 3 !}+\ldots$
$\Rightarrow \quad x=1+\frac{(1 / 2)^{\prime}}{1 !}+\frac{(1 / 2)^{2}}{2 !}+\frac{(1 / 2)^{3}}{3 !}+\ldots$
$\Rightarrow \quad x=e^{1 / 2} \Rightarrow x^{2}=e$
and $\quad y=1+\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}+\frac{x^{6}}{3 !}+\ldots$
$\Rightarrow \quad y=1+\frac{\left(x^{2}\right)^{1}}{1 !}+\frac{\left(x^{2}\right)^{2}}{2 !}+\frac{\left(x^{2}\right)^{3}}{3 !}+\ldots$
$\Rightarrow \quad y=e^{x^{2}}=e^{e} \quad$ [from Eq. (i)]
Taking log on both sides, we get
log$y=e$
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