Solving logarithmic equation $(e^x+e^{-x})/(e^x-e^{-x}) = 3$ algebraically

Question

I have the equation

$$(e^x+e^{-x})/(e^x-e^{-x}) = 3$$

I can solve it using a graph, but how would I go about solving it algebraically.

Answer 1

It can be solved by simple manipulation:

$$ e^x + e^{-x} = 3e^x - 3e^{-x} $$ $$ -2e^x + 4e^{-x} = 0 $$ $$ e^x - 2e^{-x} = 0 $$ $$ e^{2x} = 2 $$ $$ \ln{e^{2x}} = \ln{2} $$ $$ 2x = \ln{2} $$ $$ x = \dfrac{\ln{2}}{2} $$

Answer 2

If you're familiar with the hyperbolic functions, you get $$\frac{e^x+e^{-x}}{e^x-e^{-x}}=\frac{2\cosh x}{2\sinh x}=\coth x=3$$ which means $x=\coth^{-1}3$ is a solution.

Otherwise, using stochasticboy's comment, you should be able to rewrite $\coth x$ as $1+\dfrac{2}{e^{2x}-1}$.

Answer 3

Put $ e^x=t$ .

Then $\displaystyle \frac{e^x+e^{-x}}{e^x-e^{-x}}=\frac{t^2+1}{t^2-1}=3$ as $t>0$ for all $x \in \mathbb R$

For $ t \neq 1 $$\Rightarrow t^2+1=3t^2-3 \Rightarrow t^2=2 \Rightarrow e^{2x}=2$

$2x=ln(2)$.