Mix Examples-Exponential and Logarithmic Series Questions in English

(C) Given the logarithmic series expansion: $b = a - \frac{a^2}{2} + \frac{a^3}{3} - \frac{a^4}{4} + \dots = \log_e(1 + a)$.
This implies $1 + a = e^b$.
Using the exponential series expansion for $e^b$:
$e^b = 1 + \frac{b}{1!} + \frac{b^2}{2!} + \frac{b^3}{3!} + \dots$
Substituting $1 + a$ for $e^b$:
$1 + a = 1 + b + \frac{b^2}{2!} + \frac{b^3}{3!} + \dots$
Subtracting $1$ from both sides:
$a = b + \frac{b^2}{2!} + \frac{b^3}{3!} + \dots$
Thus,the value is $a$.

(B) Given equation: $e^{2x} - 11e^{x} - 45e^{-x} + \frac{81}{2} = 0$
Multiply the entire equation by $2e^{x}$:
$2e^{3x} - 22e^{2x} + 81e^{x} - 90 = 0$
Let $e^{x} = t$. Then the equation becomes:
$2t^{3} - 22t^{2} + 81t - 90 = 0$
Let the roots of this cubic equation in $t$ be $t_{1}, t_{2}, t_{3}$. These correspond to $e^{x_{1}}, e^{x_{2}}, e^{x_{3}}$.
From the properties of roots of a cubic equation $at^{3} + bt^{2} + ct + d = 0$,the product of the roots is $t_{1}t_{2}t_{3} = -\frac{d}{a}$.
Here,$t_{1}t_{2}t_{3} = -\frac{-90}{2} = 45$.
Substituting back $e^{x_{i}}$:
$e^{x_{1}} \cdot e^{x_{2}} \cdot e^{x_{3}} = 45$
$e^{x_{1} + x_{2} + x_{3}} = 45$
Taking the natural logarithm on both sides:
$x_{1} + x_{2} + x_{3} = \log_{e} 45$
Given that the sum of the roots is $\log_{e} P$,we have $\log_{e} P = \log_{e} 45$.
Therefore,$P = 45$.

(B) Given $x = \log_e 3$,then $e^x = 3$.
We know that $\tanh 2x = \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}$ and $\operatorname{sech} 2x = \frac{2}{e^{2x} + e^{-2x}}$.
Thus,$\tanh 2x + \operatorname{sech} 2x = \frac{e^{2x} - e^{-2x} + 2}{e^{2x} + e^{-2x}}$.
Since $e^x = 3$,we have $e^{2x} = (e^x)^2 = 3^2 = 9$ and $e^{-2x} = \frac{1}{9}$.
Substituting these values:
$\tanh 2x + \operatorname{sech} 2x = \frac{9 - \frac{1}{9} + 2}{9 + \frac{1}{9}} = \frac{\frac{81 - 1 + 18}{9}}{\frac{81 + 1}{9}} = \frac{98}{82} = \frac{49}{41}$.
Therefore,the correct option is $B$.