Question 755955
Having looked at this I'd do it slightly differently.

Still divide both sides by A to begin with to give

{{{V/A=1-K^(-t/T)}}}

Rather than move the 1 lets make K (the subject a positive number as it's negative atm. Add K^(-t/T) to both side and subtract V/A from both sides to give

{{{K^(-t/T)=1-(V/A)}}}

then using the rule of swapping powers, -t/T when moved to the other side of the equation reverses.

{{{K=(1-V/A)^(T/t)}}}