How to cancel an irrational power

Question

We all learned in 9th/10th grade that problems like:

$$\sqrt{x+2}=4$$

$$(x+2)^\frac 3 4=4$$

Can be solved by exponentiation

$$\text{If}: \sqrt{x+2}=4, \text{ then}:x+2=4^2$$

$$If: (x+2)^\frac 3 4=4,\text{ then}:(x+2)^3=4^4$$

And we can then expand the second problem via binomial expansion or whatever we like.

I recently encountered a problem that looked like this:

$$\frac{a(ax^2+c-d)}{a^\sqrt2}=(\frac{a^\sqrt2}{a}x^\sqrt2+d)^\sqrt2$$

(attempting to solve for d)

Is this solveable?

Before we could take advantage of the fact that the exponent was rational and so we could take any expression

$$x^\frac n m=y^\frac p q$$

and transform it to

$$x^{nq}=y^{mp}$$

No trick like this will work in solving problems with irrational exponents. Is there an alternate strategy that allows for solving problems like this?

Answer 1

You can raise things to irrational powers to clear exponents, but it doesn't help here. $$\frac{a(ax^2+c-d)}{a^\sqrt2}=(\frac{a^\sqrt2}{a}x^\sqrt2+d)^\sqrt2\\ a(ax^2+c-d)=(\frac{a^\sqrt2}{a}x^\sqrt2+d)^\sqrt2a^\sqrt2\\ a(ax^2+c-d)=((ax)^\sqrt2+ad)^\sqrt 2$$ You could raise both sides to the $\frac 1{\sqrt 2}$ power to clear the exponent on the right, but that traps the $d$ on the left.

Answer 2

Ignoring the nasty thornbush of how you define what $b^x$ can mean if $x$ is not rational and assuming we got that all worked out, we still have $(b^x)^y = b^{xy}$ if $b > 0$. (That's actually not at all trivial and stops being true for complex numbers and why this is true for reals is a significant result. But it is a question for another time.)

So for a problem like $x^{\sqrt 2} = 7$ we'd just say $(x^{\sqrt 2})^{\frac 1{\sqrt 2} } = x = 7^{\frac 1{\sqrt 2}}$.

Or we could do $(x^{\sqrt 2})^{\sqrt 2} = x^2 = 7^{\sqrt 2}$ so ($x \ge 0$ for $x^{\sqrt 2}$ to b defined) so $x = 7^{\frac {\sqrt 2}{2}}=7^{\frac 1{\sqrt 2}}$

You can do either of those on:

$\frac{a(ax^2+c-d)}{a^\sqrt2}=(\frac{a^\sqrt2}{a}x^\sqrt2+d)^\sqrt2$

$a(ax^2+c-d) = (\frac{a^\sqrt2}{a}x^\sqrt2+d)^\sqrt2 a^{\sqrt 2}$

$[a(ax^2 + c -d)]^{\sqrt 2} = ({a^\sqrt2}x^\sqrt2+da)^2$

$[a(ax^2 + c -d)]^{\frac {\sqrt 2}2} = {a^\sqrt2}x^\sqrt2+da$

which doesn't make it any easier. We'll have to come up with something else.

I'm not entirely sure what.

$\frac{a(ax^2+c-d)}{a^\sqrt2}=(\frac{a^\sqrt2}{a}x^\sqrt2+d)^\sqrt2$

$a^{1-\sqrt 2}(ax^2 + c +d) = (a^{\sqrt 2 - 1}x^\sqrt2 +d)^{\sqrt 2}$

$ax^2 + c + d = a^{\sqrt 2-1}(a^{\sqrt 2 - 1}x^\sqrt2 +d)^{\sqrt 2}$

$ax^2 + c + d = (a^{\sqrt 2-1 + \frac {\sqrt 2-1}{\sqrt 2}}x^\sqrt2 + da^{ \frac {\sqrt 2-1}{\sqrt 2}} )^{\sqrt 2}= (a^{\frac 1{\sqrt 2}}x^\sqrt2 +da^{ \frac {\sqrt 2-1}{\sqrt 2}} )^{\sqrt 2}$

Yeah, I got nothing.

In general $x^2 + k = w(x^{\sqrt{2}} + j)^{\sqrt 2}$ doesn't seem to be simplifiable.

We can transpose to $(x^2 + k)^{\frac 1 {\sqrt 2}} = w^{\frac 1{\sqrt 2}}(x^\sqrt{2} + j)$

or even $x = (w(x^{\sqrt{2}} + j)^{\sqrt 2} -k)^{\frac 12}$ but... I just don't see how we can distribute through powers.