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I'm having some problems in obtaining $x$ from this example $y=0,1001\cdot \ln(x)+0,3691$. After some aptempts I got: $$e^{((y-0,3691)/0,1001)}=x$$ Is this even close to the correct result?

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    Yes perfectly fine.2017-01-25

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Since you already have the correct solution, I think it's valid to just give you the necessary steps to obtain it, so you can check your calculation. You started with:

$y = 0.1001 \cdot ln(x) + 0.3691$

You subtract $0.3691$ from both sides, then you divide by $0.1001$ on both sides. You end up with:

$\frac{y-0.3691}{0.1001} = ln(x)$.

Now you need the knowledge that the inverse function of $e^x$ is $ln(x)$, in other words, $ln(e^x)=x$ and also $e^{ln(x)}=x$. So you can take the values on both sides and put them in the exponent of the exponential function and end up with:

$e^{\frac{y-0.3691}{0.1001}} = e^{ln(x)} = x$.

In other words, we just showed that

$x= e^{\frac{y-0.3691}{0.1001}}$.

And this is the result you managed to obtain yourself, so I think you followed a similar chain of steps as well.