Move variable to one side of the equation

Question

Say if I have a formula like so:

a1*a2*a3^(a4 + 1)*(1 - E^(a5*a6/a3^a4/a2)) == 0

How do I move a3 to the right? I've tried to follow other examples here on stack exchange but couldn't find anything that worked on this kind of formula.

Using Solve[] causes this error:

Solve::nsmet: This system cannot be solved with the methods available to Solve.

Answer 1

I assume you mean by move a3 to the right that you want to solve for a3. Are you working in real numbers, then Reduce might be a better option than Solve

Reduce[a1*a2*a3^(a4 + 1)*(1 - E^(a5*a6/a3^a4/a2)) == 0, a3, Reals]

and you get

(-1 < a4 < 0 && a3 == 0) || (C[1] \[Element] Integers && (
(a1 == 0 && a4 == -C[1] && a3 < 0) || (a2 == 0 && 
a4 == -C[1] && a3 < 0) || (a5 == 0 && a4 == -C[1] && 
a3 < 0) || (a6 == 0 && a4 == -C[1] && a3 < 0))) || (a1 == 0 && 
a3 > 0) || (a2 == 0 && a3 > 0) || (a5 == 0 && a3 > 0) || (a6 == 0 && a3 > 0)

Now you have to understand the solution. Look at the structure of your equation. It's a product and therefore it is zero when either factor is zero. All the possibilities given (as logical formula) tell you how it is possible to make one factor zero. For instance setting a3==0 but this holds only, if the exponent containing a4 fulfills some requirements, namely -1 < a4 < 0.

If you are really only interested in the solution that contains a3, you can use Solve which gives you effectively only the first part of the solution of Reduce

Solve[a1*a2*a3^(a4 + 1)*(1 - E^(a5*a6/a3^a4/a2)) == 0, a3, Reals]

Out[4]= {{a3 -> ConditionalExpression[0, -1 < a4 < 0]}}