The solution of the equation 2 cosh 2x + 10 s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the equation: $2 \cosh 2x + 10 \sinh 2x = 5$ Substitute the exponential definitions $\cosh 2x = \frac{e^{2x} + e^{-2x}}{2}$ and $\sinh 2x =

Vedclass · Accepted Answer

$\frac{1}{2} \log \left(\frac{4}{3}\right)$

Vedclass · Answer

(B) Given the equation: $2 \cosh 2x + 10 \sinh 2x = 5$ Substitute the exponential definitions $\cosh 2x = \frac{e^{2x} + e^{-2x}}{2}$ and $\sinh 2x = \frac{e^{2x} - e^{-2x}}{2}$: $2 \left(\frac{e^{2x} + e^{-2x}}{2}\right) + 10 \left(\frac{e^{2x} - e^{-2x}}{2}\right) = 5$ $(e^{2x} + e^{-2x}) + 5(e^{2x} - e^{-2x}) = 5$ $e^{2x} + e^{-2x} + 5e^{2x} - 5e^{-2x} = 5$ $6e^{2x} - 4e^{-2x} = 5$ Multiply by $e^{2x}$: $6(e^{2x})^2 - 5e^{2x} - 4 = 0$ Let $u = e^{2x}$,then $6u^2 - 5u - 4 = 0$ Factor the quadratic: $6u^2 - 8u + 3u - 4 = 0 \Rightarrow 2u(3u - 4) + 1(3u - 4) = 0$ $(2u + 1)(3u - 4) = 0$ Since $u = e^{2x} > 0$,we must have $3u - 4 = 0$,so $u = \frac{4}{3}$ $e^{2x} = \frac{4}{3} \Rightarrow 2x = \log \left(\frac{4}{3}\right)$ $x = \frac{1}{2} \log \left(\frac{4}{3}\right)$

The solution of the equation $2 \cosh 2x + 10 \sinh 2x = 5$ is

Similar Questions

$\frac{x^2 - y^2}{1!} + \frac{x^4 - y^4}{2!} + \frac{x^6 - y^6}{3!} + \dots \infty = $

The sum of the series $\frac{2}{2 !} + \frac{2+4}{3 !} + \frac{2+4+6}{4 !} + \ldots$ is equal to

If ${T_n} = \frac{{{3^n}}}{{2(n!)}} - \frac{1}{{2(n!)}}$,then ${S_\infty } = $

$\frac{1^2 \cdot 2}{1!} + \frac{2^2 \cdot 3}{2!} + \frac{3^2 \cdot 4}{3!} + \dots \infty = $ (in $e$)

$\frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \frac{8}{7!} + \dots \infty = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module