# Solve for a constant in an equation

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1. Jan 14, 2016

### Dean Whaley

Im trying to solve for a constant in an equation and it involves taking the arctanh(6.55) and my calculator is giving me an error, is there a way around this?

2. Jan 14, 2016

### H Smith 94

Perhaps you could make use of $$\tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^x - e^{-x} }{ e^x + e^{-x} },$$ or $$\mathrm{arctanh}(x) = \frac{1}{2}\,\ln\left|\frac{1+x}{1-x} \right|.$$ These are all available by Googling "Hyperbolic trig identities".

Also, you should probably mention what the equation is.

Last edited: Jan 14, 2016
3. Jan 15, 2016

### Dean Whaley

This helps, thanks a lot

4. Jan 15, 2016

### nasu

The values of tanh are between -1 and 1.
What you are trying to do is like asking for arcsin(2).
Of course you get an error.:)

5. Jan 15, 2016

### Staff: Mentor

If you let x = arctanh(6.55), an equivalent equation is tanh(x) = 6.55.

Assuming that we're dealing only with real numbers, the range of the tanh function is $-1 < \tanh(x) < 1$. This means that there is no real number x for which tanh(x) = 6.55, or equivalently, for which x = arctanh(6.55). That's what your calculator is telling you.

Your calculator probably has some documentation about the values that can be used as arguments to each of the calculator's functions.

Edit: nasu beat me by 2 minutes!