$\begingroup$If I know the outcome of a value being raised to a certain power, is it possible to know what the original base was?
Example:
x ^ 0.25 = 2.5045 What's the proper way to calculate x?
$\endgroup$23 Answers
$\begingroup$The proper way to do $x^b = a$ is to both sides to $1/b$.
$x = (x^b)^{1/b} = a^{1/b}$.
So $x = (x^{.25})^{1/.25} = 2.5045^{1/.25} = 2.5045^4 = 39.3445102866600625$.
$\endgroup$$\begingroup$They way to generally do this would be that you take the logarithm of both sides of the equation. Then you would have:
$$ \log(x ^{0,25} ) = \log(2.5045) $$
Then, since $$ \log(a^b) = b * \log(a) $$
you end up with $$ 0,25 * \log(x) = \log(2.5045) $$
which implies that $$ 10^{(\log(2.5045) / 0,25)} = x $$ by the definition of the logarithm
$\endgroup$5$\begingroup$In this particular example you can rewrite $x^{0.25}=x^{\frac{1}{4}}$
Then you can just put $x=2,5045^{\frac{4}{1}}$
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