Aubrie Mccall

2022-10-07

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

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

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

Tanya Anthony

Beginner2022-10-08Added 5 answers

It can be solved by simple manipulation:

$${e}^{x}+{e}^{-x}=3{e}^{x}-3{e}^{-x}$$

$$-2{e}^{x}+4{e}^{-x}=0$$

$${e}^{x}-2{e}^{-x}=0$$

$${e}^{2x}=2$$

$$\mathrm{ln}{e}^{2x}=\mathrm{ln}2$$

$$2x=\mathrm{ln}2$$

$$x={\displaystyle \frac{\mathrm{ln}2}{2}}$$

$${e}^{x}+{e}^{-x}=3{e}^{x}-3{e}^{-x}$$

$$-2{e}^{x}+4{e}^{-x}=0$$

$${e}^{x}-2{e}^{-x}=0$$

$${e}^{2x}=2$$

$$\mathrm{ln}{e}^{2x}=\mathrm{ln}2$$

$$2x=\mathrm{ln}2$$

$$x={\displaystyle \frac{\mathrm{ln}2}{2}}$$

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