Question

How do you solve $13\sinh x-7\cosh x+1=0$ ?

Answer 1

These types of problems are pretty straight forward and are very simple to solve. To solve these types of problems, we need to have an in-depth idea of hyperbolic functions as well as their related equations. We substitute their respective values and then evaluate for the required answer. The above given problem requires us to find the value of ‘x’, which we can pretty easily do after we have substituted the values in the form of equations in place of $\sinh x$ and $\cosh x$ . Some of the equations related to hyperbolic functions are as follows,

& \sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \\\ & \cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \\\ \end{aligned}$$ **Complete step by step solution:** Now we start off with our solution by replacing the formulae for the hyperbolic functions in the given problem, we can thus write, $$13\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}-7\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}+1=0$$ Now, further simplifying down the equation, we can write it as, $$\Rightarrow \dfrac{13}{2}{{e}^{x}}-\dfrac{7}{2}{{e}^{x}}-\dfrac{13}{2}{{e}^{-x}}-\dfrac{7}{2}{{e}^{-x}}+1=0$$ Now doing the necessary additions and subtractions, we get, $$\Rightarrow 3{{e}^{x}}-10{{e}^{-x}}+1=0$$ Now, dividing by $${{e}^{-x}}$$ throughout, we get, $$\Rightarrow 3{{e}^{2x}}+{{e}^{x}}-10=0$$ This basically forms a quadratic equations, and we can find its roots as, $$\begin{aligned} & {{e}^{x}}=\dfrac{-1\pm \sqrt{{{1}^{2}}-4.3.\left( -10 \right)}}{2.3} \\\ & \Rightarrow {{e}^{x}}=\dfrac{-1\pm \sqrt{121}}{2.3} \\\ & \Rightarrow {{e}^{x}}=\dfrac{-1\pm 11}{2.3} \\\ & \Rightarrow {{e}^{x}}=\dfrac{-1+11}{6},\dfrac{-1-11}{6} \\\ & \Rightarrow {{e}^{x}}=\dfrac{5}{3},-2 \\\ \end{aligned}$$ Now, taking logarithm on both the sides of the equation, we find the value of ‘x’ as, $$\begin{aligned} & \Rightarrow {{e}^{x}}=\dfrac{5}{3},-2 \\\ & \Rightarrow x=\ln \left( \dfrac{5}{3} \right) \\\ \end{aligned}$$ Since logarithm of a negative number is not possible, so we eliminate the $$-2$$ term and find out the answer for the left out term, Thus, $$\begin{aligned} & x=\ln \left( \dfrac{5}{3} \right) \\\ & \Rightarrow x=0.51083 \\\ \end{aligned}$$ **Note:** For problems like these, we need to be very thorough with the theory of hyperbolic functions and their related equations. We must also have a fair idea of logarithms and antilogarithms which might be useful at times. We must also remember the graph for a logarithmic function and know that the logarithmic for a negative number doesn’t exist, so only the positive value is taken into account.