2 times 3 to the 2x minus 5 times 3 to the x plus 3=0

Question

It is in word form above but numerically it is ((2*3)^2x)-((5+3)^x)+3=0 solve for x.

Answer 1

6^2x-8^x+3=0 has no (real) solutions.

Perhaps the question should be:

2(3^2x)-(5+3^x)+3=0. Let y=3^x, then:

2y²-5-y+3=0,

2y²-y-2=0,

y²-y/2=1,

y²-y/2+1/16=17/16,

(y-¼)²=17/16,

y-¼=√17/4, y=¼±¼√17, y=1.28078 or -0.78078 approx.

Therefore 3^x=1.28078 (the negative y is rejected because no power of 3 has a negative result).

xlog(3)=log(1.28078), x=log(1.28078)/log(3)=0.22525 approx.

Or is the question 2(3^2x)-5(3^x)+3=0?

Let y=3^x, then 2y²-5y+3=0=(y-1)(2y-3), so y=1 or 1.5.

3^x=1, so x=0; or 3^x=1.5, x=log(1.5)/log(3)=0.3691 approx.