How do I calculate the base of an exponent if I know the result and what it was raised to?

Question

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?

Raise it to the reciprical power: $x^n = b$ so $x = (x^n)^{1/n} = b^{1/n}$. So $x^{1/4} =2.5045 \implies $x = 2.5045^4$. — fleablood, Oct 10 '16 at 16:08

score 7 · Answer 1 · answered Oct 10 '16 at 16:16

7

The proper way to do $x^b = a$ is to both sides to $1/b$.

$x = (x^b)^{1/b} = a^{1/b}$.

So $x = (x^{.25})^{1/.25} = 2.5045^{1/.25} = 2.5045^4 = 39.3445102866600625$.

answered Oct 10 '16 at 16:16

fleablood

score 0 · Answer 2 · answered Oct 10 '16 at 15:59

0

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

answered Oct 10 '16 at 15:59

AxiomaticApproach

1

But $10^{\log(2.5045)/0.25} = 2.5045^{1/.25} = 2.5045^4$. Seeing as we don't know what $x$ is taking the log of x doesn't get us closer and this seems a very round about and obscure way to simply raise 2.5045 to the 4th power. – fleablood Oct 10 '16 at 16:12
Good point. I guess its just overly complicated. – AxiomaticApproach Oct 10 '16 at 19:01
No. It's always best to isolate the variable. If the variable has a number added to it, then you subtract. If the variable is multiplied by a number then you divide. If the variable is raised to a power then you raise it to the reciprical power. If the variable it a base raised to the variable power THEN you take a log. You always "strip" the variable by doing the inverse. If you have $b^x$ then you take a log. But if you have $x^b$ taking a log will get you nowhere. $x^{.123489} = K$ is solved by ..... – fleablood Oct 10 '16 at 19:05
Yep, makes sense – AxiomaticApproach Oct 10 '16 at 19:08
$x = K ^{1/.123489}$. Taking log, $.123489 \log x = log K$ so $\log x = \log K^{1/.123489}$ so $x = 10^{\log K^{1/.123489}} = k^{1/.123489}$. SO taking log was just an indirect waste of time. – fleablood Oct 10 '16 at 19:08

score 0 · Answer 3 · answered Oct 10 '16 at 16:05

0

In this particular example you can rewrite $x^{0.25}=x^{\frac{1}{4}}$

Then you can just put $x=2,5045^{\frac{4}{1}}$

answered Oct 10 '16 at 16:05

Arne Morten

3 Answers3