2023/030) Solve in real x $6^x + 9^x =2^(2x+1)$

As we see above power of 2 and 3 ( 9 is $3^2$) and 6 are invloved

Let $3^x = a$ and $2^x=b$

We get $ab + a^2 = 2 b^2$

Or $a^2 + ab - 2b^2 = 0$

 or $(a-b)(a+2b) = 0$

$a=b$ or $a+2b=0$

as a and b are  positive s a = b (a+2b=0 is inadmissible)

or $3^x = 2^x$ or x = 0