Solving for a variable that's an index of e

Question

I have arrived at the equation

$$0=e^{10b}+e^{-10b}-2.5$$

and was wondering how I would go about calculating $b$.

I was considering using the quadratic formula but I wasn't sure if it would work since one index isn't raised higher than the other.

Answer 1

HINT:

As $e^{10b}$ is non-zero finite, multiply both sides by $e^{10b}$ to get $$(e^{10b})^2-2.5(e^{10b})+1=0$$ which is a Quadratic Equation in $e^{10b}$

Now solve and take log with respect to $e$

Answer 2

We start off knowing that $$e^{10b}+e^{-10b}=2.5$$ $$\frac{e^{10b}+e^{-10b}}{2}=1.25$$ Now we can use the "$\cosh$" function, or the hyperbolic cosine function (or rather, its inverse). Just as a reminder, the hyperbolic cosine function is $$\cosh x=\frac{e^x+e^{-x}}{2}$$ and its inverse is $$\cosh^{-1}x=\ln(x\pm\sqrt{x^2-1})$$ So if we apply $\cosh^{-1}$ to both sides of your equation, we get $$10b=\ln(1.25\pm\sqrt{1.25^2-1})$$ $$10b=\ln(1.25\pm0.75)$$ $$b=\frac{\ln(1.25\pm0.75)}{10}$$ $$b=\pm\frac{\ln(2)}{10}$$

Answer 3

You can substitute. $0=e^{10b}+e^{-10b}-2.5$

Replace $e^{10b}$ by $x$.

$$0=x+\frac1x-2.5$$

Multplying the equation by x.

$x^2+1-2.5x=0$

$x^2-2.5x+1=0$

Using the quadratic formula to solve it.

$x_{1/2}=\frac{2.5\pm \sqrt{6.25-4}}{2}$

$x_{1/2}=\frac{2.5\pm 1.5}{2}$

$x_1=2, x_2=0.5$

Re-substitution

$$2=e^{10b_1}\Rightarrow \ln(2)=10b_1 \Rightarrow b_1=\frac{\ln(2)}{10}$$

$$0.5=e^{10b_2}\Rightarrow \ln(0.5)=10b_2 \Rightarrow b_2=\frac{-\ln(2)}{10}$$

Answer 4

We have $$e^{10b}+e^{-10b}-2.5=0$$

We can multiply both sides by $e^{10b}$ to get

\begin{align}e^{10b}\left(e^{10b}+e^{-10b}-2.5\right)&=e^{10b}\times 0\\ \left(e^{10b}\right)^2+1-2.5e^{10b}&=0\\ \left(e^{10b}\right)^2-2.5e^{10b}+1&=0\end{align}

We use the substitution $t=e^{10b}$ and rewrite this as \begin{align}t^2-2.5t+1&=0\\ \left(t-\frac{2.5}{2}\right)^2-\left(\frac{2.5}{2}\right)^2+1&=0\\ (t-1.25)^2-1.5625+1&=0\\ (t-1.25)^2-0.5625&=0\\ (t-1.25)^2&=0.5625\\ t-1.25&=\pm\sqrt{0.5625}\\ t-1.25&=\pm0.75\\ t&=1.25\pm0.75\\ t&=\{2,0.5\}\end{align}

So now we have to solve $e^{10b}=t$

Firstly, we have \begin{align}e^{10b}&=2\\ \ln\left(e^{10b}\right)&=\ln(2)\\ 10b&=\ln(2)\\ b&=\frac{\ln(2)}{10}\end{align}

We also have \begin{align}e^{10b}&=\frac 12\\ \ln\left(e^{10b}\right)&=\ln\left(\frac 12\right)\\ 10b&=\ln(1)-\ln(2)\\ 10b&=-\ln(2)\\ b&=-\frac{\ln(2)}{10}\end{align}

Therefore we have our answer: $$b=\pm\frac{\ln(2)}{10}$$

---

Note: I am using the following logarithm laws:

\begin{align}\ln\left(e^a\right)&=a\\ \ln\left(\frac ab\right)&=\ln(a)-\ln(b)\end{align}