# Find all solutions of the equation 2^(x^2-5x+10)-64=0

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The equation to be solved is: 2^(x^2 - 5x + 10) - 64 = 0

2^(x^2 - 5x + 10) - 64 = 0

=> 2^(x^2 - 5x + 10) = 64

64 = 2^6

=> 2^(x^2 - 5x + 10) = 2^6

as the base is the same equate the exponent

x^2 - 5x + 10 = 6

=> x^2 - 5x + 4 = 0

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

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

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

=> x = 1 and x = 4

**The solution of the equation is (1 , 4)**

The equation to be solved is 2^(x^2-5x+10)-64=0

This can be written as 2^(x^2-5x+10) = 64

Note that 64 = 2^6

2^(x^2-5x+10) = 64 = 2^6

As the base is the same, equate the exponent

x^2 - 5x + 10 = 6

x^2 - 5x + 4 = 0

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

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

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

x = 1 and x = 4

The solution of the equation is x = 1 and x = 4

We'll shift the number alone to the right side:

2^(x^2-5x+10) = 64

We'll create matching bases both sides. For this reason, we'll re-write 64 = 2^6

We'll re-write the equation as it follows:

2^(x^2-5x+10)= 2^6

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

(x^2-5x+10)=6

We'll subtract 6 both sides:

x^2-5x+10-6=0

x^2-5x+4=0

We'll apply quadratic formula:

x1=[5+sqrt(25-16)]/2

x1=(5+3)/2

x1=4

x2=(5-3)/2

x2=1

**The complete set of solutions of the exponential equation is: {1 ; 4}.**