3*3^2x-7*12^x+4*4^2x=0

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To solve the given equation, we can simplify it step by step.

33^(2x) - 712^x + 4*4^(2x) = 0

Rewrite the terms using exponents:

3(3^x)^2 - 7(2^2 3^x) + 4(2^2)^x = 0

Now, let's simplify the terms further:

3(3^x)^2 = 3 9^x = 27^x
7(2^2 3^x) = 283^x
4(2^2)^x = 44^(2x) = 4 16^x

Substitute these values back into the original equation:

27^x - 283^x + 416^x = 0

Now, we have an equation with terms in a similar form. Let's rewrite it with a single base:

3^x = y

27^x = (3^3)^x = 3^(3x) = y^3
16^x = (4^2)^x = 4^(2x) = (2^2)^2x = 2^4x = (2^x)^4 = y^4

With these replacements, our equation becomes:

y^3 - 28y + 4y^4 = 0

Now, this equation is a quadratic equation in terms of y. Let's solve it to find the possible values of y, and then we can go back to find x.

0 = 4y^4 + y^3 - 28y

Let's factor this equation by factoring out y:

0 = y(4y^3 + y^2 - 28)

Now, let's factor the quadratic expression inside the parentheses:

0 = y(4y^2 - 7y + 4)

Factor the quadratic equation:

0 = y(4y^2 - 7y + 4)
0 = y(4y^2 - 2y - 5y + 4)
0 = y(2y(2y - 1) - 1(2y - 1))
0 = y(2y - 1)(2y - 1)

Now, we have three possible solutions for y:

y = 0, y = 1/2, y = 1/2

Now, substitute back the values of y into 3^x = y:

Case 1: y = 0
3^x = 0
This is not possible because 3 raised to any power will never equal 0.

Case 2: y = 1/2
3^x = 1/2
Take the logarithm of both sides:
x*log(3) = log(1/2)
x = log(1/2)/log(3)

Case 3: y = 1
3^x = 1
x = 0

To summarize, the possible solutions to the original equation are x = log(1/2)/log(3) or x = 0.