**Solving the Equation 64***9^x-84*12^x+27*16^x=0

*9^x-84*12^x+27*16^x=0

In this article, we will explore the solution to the equation 64*9^x-84*12^x+27*16^x=0. This equation involves exponential functions and requires some clever manipulation to solve.

**Rewriting the Equation**

Let's start by rewriting the equation to make it more manageable:

64(3^2)^x - 84(3^2 * 2^2)^x + 27(2^4)^x = 0

Simplifying the equation, we get:

64(3^2)^x - 84(3^2)^x(2^2)^x + 27(2^4)^x = 0

**Factoring Out Common Terms**

Notice that (3^2)^x is a common term in the first two expressions. Let's factor it out:

(3^2)^x [64 - 84(2^2)^x] + 27(2^4)^x = 0

Simplifying further, we get:

(3^2)^x [64 - 84(4)^x] + 27(2^4)^x = 0

**Substituting y = 2^x**

To simplify the equation, let's substitute y = 2^x:

(3^2)^(x/2) [64 - 84y^2] + 27y^4 = 0

Now we have a polynomial equation in y:

27y^4 - 84y^2 + 64 = 0

**Factoring the Quadratic**

This is a quadratic equation in y^2. Factoring it, we get:

(3y^2 - 4)(9y^2 - 16) = 0

**Solving for y**

Now we have two factors. Setting each factor equal to 0, we get:

3y^2 - 4 = 0 --> y^2 = 4/3 --> y = ±2/\√3

9y^2 - 16 = 0 --> y^2 = 16/9 --> y = ±4/3

**Solving for x**

Now that we have the values of y, we can solve for x:

y = 2^x --> 2^x = ±2/\√3 or 2^x = ±4/3

Taking the logarithm base 2 of both sides, we get:

x = log2(±2/\√3) or x = log2(±4/3)

Simplifying, we get:

x = log2(2/\√3) or x = log2(4/3)

x = 1 - log2(\√3) or x = 2 - log2(3)

**Conclusion**

In conclusion, we have solved the equation 64*9^x-84*12^x+27*16^x=0 and found the solutions for x. The solutions involve logarithmic functions, which may require numerical methods to evaluate.