How to cancel an irrational power 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? 

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.
