Congratulations@goodness barak

@Goodness Barak Got it and also finalized the solution. So the credit goes to him.

@Harbdul Qudus got it on the second attempt but you did not finish the question. Log format is a question on its own.

The equation can be expressed as
2^2x +(2×3)^x=(3²)^x
2^2x + 2^x × 3^x=3^2x
2^2x + 2^x × 3^x - 3^2x = 0
Let 2^x=a ,3^x=b
Substituting...
a² + ab + b² =0
Dividing through by b²
(a²/b²) + (ab/b²) - (b²/b²)=0
b² cancel out b² , b cancel out one b from b² we now have....
(a/b)² + (a/b) -1= 0
To make the expression simpler, Let a/b = c .... We have...
c² + c - 1=0
Solving using quadratic formula...
c= 0.6180 or -1.6180
Since c= a/b =2^x/3^x which is an exponential function which implies that c cannot be negative so we'll be using the positive answer...
c= 0.6180
Since c = (a/b) =2^x/3^x = (2/3)^x , we can now substitute for c....
(2/3)^x = 0.6180
Taking the log of both sides ...
log(2/3)^x = log 0.6180
xlog(2/3) = log 0.6180
Dividing both sides by log (2/3)
x= (log 0.6180) / (log 2/3)
x = (-0.2090) / (-0.1760)
x= 1.1868

The equation can be expressed as
2^2x +(2×3)^x=(3²)^x
2^2x + 2^x × 3^x=3^2x
2^2x + 2^x × 3^x - 3^2x = 0
Let 2^x=a ,3^x=b
Substituting...
a² + ab + b² =0
Dividing through by b²
(a²/b²) + (ab/b²) - (b²/b²)=0
b² cancel out b² , b cancel out one b from b² we now have....
(a/b)² + (a/b) -1= 0
To make the expression simpler, Let a/b = c .... We have...
c² + c - 1=0
Solving using quadratic formula...
c= 0.6180 or -1.6180
Since c= a/b =2^x/3^x which is an exponential function, c cannot be negative so we'll be using the positive answer...
c= 0.6180
Since c = (a/b) =2^x/3^x = (2/3)^x , we can now substitute for c....
(2/3)^x = 0.6180
Taking the log of both sides ...
log(2/3)^x = log 0.6180
xlog(2/3) = log 0.6180
Dividing both sides by log (2/3)
x= (log 0.6180) / (log 2/3)
x = (-0.2090) / (-0.1760)
x= 1.1868