Question

Sinhx=5 prove that x= loge(5+√26

Answer 1

$\textbf{Given:}$

$\mathrm{sinhx=5}$

$\textbf{To prove:}$

$\mathrm{x=log_e(5+\sqrt{26})}$

$\textbf{Solution:}$

$\text{Consider,}$

$\mathrm{sinhx=5}$

$\mathrm{\dfrac{e^x-e^{-x}}{2}=5}$

$\mathrm{e^x-e^{-x}=10}$

$\mathrm{e^x-\dfrac{1}{e^x}=10}$

$\mathrm{(e^x)^2-1=10\,e^x}$

$\mathrm{(e^x)^2-10\,e^x-1=0}$

$\mathrm{t^2-10t-1=0}\;\;\text{where}\,\mathrm{t=e^x}$

$\text{Using quadratic formula, we get}$

$\mathrm{t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}$

$\mathrm{t=\dfrac{10\pm\sqrt{100+4}}{2}}$

$\mathrm{t=\dfrac{10\pm\sqrt{104}}{2}}$

$\mathrm{t=\dfrac{10\pm\sqrt{4{\times}26}}{2}}$

$\mathrm{t=\dfrac{10{\pm}2\sqrt{26}}{2}}$

$\mathrm{t=5\pm\sqrt{26}}$

$\implies\mathrm{e^x=5\pm\sqrt{26}}$

$\implies\mathrm{x=log_e(5\pm\sqrt{26})}$

$\text{But x cannot be negative}$

$\boxed{\bf\mathrm{x=log_e(5+\sqrt{26})}}$

$\textbf{Hence proved}$