Let (P,Q)= (2^x, 3^x)

At the outset, we observe that each of P and Q must be non zero. Otherwise, each of 2^x and 3^equals 1 which does not have any solution.

Now, the transformed equation due to the ano e substitution is given by:

P+Q-P^2+PQ-Q^2=0

=>2P+2Q-2P^2+2PQ-2Q^2=0 (multiplying both sides by 2)

=> 2P^2-2PQ+2Q^2-2P-2Q=0

=> P^2-2P+Q^2-2Q+P^2+Q^2-2PQ=0

=> (P-1)^2+(Q-1)^2+(P-Q)^2=2

Since lhs this a sum of three square, it follows that precisely two of the terms must equal 1. and the third must equal 0

CASE 1: ((P-1)^2=0

=>P-1=0

=>P=1,

=>2^x=1

=>x=0

.

CASE 2: (Q-1)^2=0

In a similar vain as in Case 1, we get 3^x=1, giving:x=0

CASE 3:(P-Q)^2=0

Here, P-Q=0, so that: P=Q

Thus, 2^x=3^x

= > (2/3)^x=1

=>x=0

Consequently, we get x=0 as the only possible real solution to the given equation under reference.

*** It may be observed that NO GRAPHING was utilized for solving this problem.

**** Also, I made full use of the hint "Multiply by 2" given in the title itself.

*Edited on ***January 7, 2022, 3:22 am**