real solution

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If we write as (t²)^x/2+(v²)^x/2 = (t²+v²)^x/2 and set x/2 = y,

the equation becomes

a^y+b^y=(a+b)^y

Dividing by b^y, we get

1+k^y = (1+k)^y where 0< k = a/b<1

We will prove that equality occurs only when y = 1

Suppose y<0, then LHS>1 but RHS<1 so equality cannot occur

Consider f(y) = (1+k)^y-k^y -1

For y>0, (1+k)^y increases monotonically.
k^y decreases monotonically (k<1), so that -k^y increases monotonically

Now we can therefore infer that (1+k)^y-k^y -1 is a monotnically increasing continuous function

f(0) = -1, f(1) = 0. Thus f(y)<0 for 0≤y<1 and f(y)>0 for y>1

Hence there are no solutions other than y = x/2 = 1. Hence x =2 is the only solution