Given the equation, of a group A. Determine x in terms of j,k,l.
$$k∗x∗x = j∗l$$
Given that we know that its a group, we know that the binary operation has an identity, each element has an inverse and it is associative. We can add $k^{-1}$ to the left of each side by using an operation. That is,
$$k^{-1} ∗ k ∗ x ∗ x = k^{-1} ∗ j ∗ l$$
We can use the associative property to show that
$$k^{-1} ∗ k ∗ x ∗ x = (k^{-1} ∗ k) ∗ (x ∗ x)$$
And since k^{-1} ∗ k = e (From the inverse property). We get,
$$e ∗ x ∗ x = k^{-1} ∗ j ∗ l$$
and since e ∗ x = x (From the identity property). We get,
$$x ∗ x = k^{-1} ∗ j ∗ l$$
This is where I end up but I can't determine x in terms of j,k,l without having an inverse term alongside it.
Either I get
$$x = x^{-1} ∗ k^{-1} ∗ j ∗ l$$, or I get $$x = k^{-1} ∗ j ∗ l ∗ x^{-1}$$
Is it possible to determine x in terms of j,k,l for this example? And if so, where should I start/what have I done incorrectly?