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If I have $$ \mathrm{asinh}\left(\frac{x}{2.8\cdot10^{-10}}\right) = 15 $$ How can I calculate $x$? Should I use $\mathrm{asinh} \, x = \ln(x+\sqrt{x^2+1})$ Or something else?

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  • $\begingroup$ Why not apply $\sinh$ to both sides? $\endgroup$ – Arthur Mar 7 at 13:46
  • $\begingroup$ (x/2.8*10^-10) = ASINH(15.0) you mean that? $\endgroup$ – JESUS_M Mar 7 at 13:49
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    $\begingroup$ Let's say this, then. If you had the eqution $\operatorname{asinh}(y) = 15$, how would you solve it? Can you think of a single step that solves that equation? (Think about what $\operatorname{asinh}$ means. Not its formula, but its meaning / definition.) $\endgroup$ – Arthur Mar 7 at 13:50
  • $\begingroup$ asinh(y) = ln(y +sqrt(y^2 + 1) = 15 then ln(y +sqrt(y^2 + 1) = 15 and I can calculate y $\endgroup$ – JESUS_M Mar 7 at 13:59
  • $\begingroup$ I said not to use the formula. What is the main reason for $\operatorname{asinh}$ to exist? What is its purpose? $\endgroup$ – Arthur Mar 7 at 14:00
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Well, we're trying to solve:

$$\ln\left(\frac{x}{\frac{14}{5}\cdot10^{-10}}+\sqrt{1+\left(\frac{x}{\frac{14}{5}\cdot10^{-10}}\right)^2}\right)=15\space\Longleftrightarrow\space x=\frac{7\sinh\left(15\right)}{25000000000}\tag1$$

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