# What are the solutions of the equation 4*2^(x^2)/2^(3x)=64?

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The equation that we have to find the solutions of is 4*2^(x^2)/2^(3x)=64

4*2^(x^2)/2^(3x)=64

=> 2^(x^2)/2^(3x)=64/4

=> 2^(x^2)/2^(3x)=16

=> 2^(x^2) = 16*2^(3x)

=> 2^(x^2) = 2^4*2^(3x)

=> 2^(x^2) = 2^(4 + 3x)

As the base is the same equate the exponent

x^2 = 4 + 3x

=> x^2 - 3x - 4 = 0

=> x^2 - 4x + x - 4 = 0

=> x(x - 4) + 1(x - 4) = 0

=> (x + 1)(x - 4) = 0

This gives x = -1 and x = 4

**The solutions of the equation are x = -1 and x = 4**

We'll divide both sides by 4:

2^(x^2)/2^(3x) = 16

We'll re-write the equation, using quotient property of exponentials, in this way:

2^(x^2-3x)= 2^4

Since the bases are matching now, we'll apply one to one rule and we'll get:

(x^2-3x) = 4

We'll subtract 4 both sides:

x^2 - 3x - 4 = 0

We'll apply quadratic formula:

x1=[(-3)+sqrt(9+16)]/2

x1=(3+5)/2

x1=4

x2=[(-3)-sqrt(9+16)]/2

x2=(3-5)/2

x2=-1

**The real solutions of the given exponential equation are: {-1 ; 4}.**