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Sorry for the simple question, but it is so simple that I cannot seem to find the answer!

At school, I was always taught that i can use this identity:

$$a = e^{ln(a)} $$

To rewrite more complex expressions, often in order to get rid of an exponent. For example:

$$\sqrt[4]{(x+1)} = (x+1)^{1/4} = e^{ln[(x+1)^{1/4}]} = e^{\frac{1}{4}ln(x+1)}$$

But nobody ever told me that this can be done if, and only if, the codomain of the original function (the one that we are going to rewrite as an exponential) is always greater than zero. Am I correct in my assumption?

For example, if I were to write:

$$ x^3= e^{3ln(x)}$$

I would be making a mistake, since the last exponential expression $e^{3ln(x)}$ is NOT equal to the first one, $x^3$, because $x^3$ can yield negative values (when $x$ is negative) whereas the codomain of $e^{3ln(x)}$ (or even $e^{3ln(|x|)}$, if we wanted to use the absolute value) is $\mathbb{R^+}$.

Can you please provide any insight to this?

Thanks in advance

Gamma Sigma
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    Well you're right, (except that it can't be 0, either). The natural logarithm is only defined for positive values. I don't know what more insight to give you -- you seem to understand the matter perfectly, except for that small slip about 0. – saulspatz Jan 20 '18 at 12:08
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    A similar problem applies to even simpler formulae. Solve $x^2 = 2$. You can reach for your calculator and use the $\sqrt{}$ button but you will only get one of the solutions. You need to add some mathematical knowledge to find all of the solutions. It can be worse, solve: $\sin(x) = 1$. Your calculator will give one but there are infinitely many solutions. – badjohn Jan 20 '18 at 12:11
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    It's a pity that noone ever told you that ... – Hagen von Eitzen Jan 20 '18 at 12:17
  • Thank you for the comments.

    I have corrected the slip about the domain of the logarithm being non-negative instead of strictly positive.

     – Gamma Sigma Jan 21 '18 at 14:10

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